RS Aggarwal Solutions for Class 12 Chapter 21 Linear Differential Equations

Question 1. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + \frac{1}{x}.y = x^2 \)
Answer: The given differential equation is \( \frac{dy}{dx} + \frac{1}{x}.y = x^2 \).

This matches the standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x} \) and \( Q = x^2 \).

The integrating factor (I.F.) is calculated as:

\( I.F. = e^{\int P \, dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x \)

Using the general solution formula:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot x = \int x^2 \cdot x \, dx + c \)

\( xy = \int x^3 \, dx + c \)

\( xy = \frac{x^4}{4} + c \)

\( y = \frac{x^3}{4} + \frac{c}{x} \)

Exam Tip: Always identify P and Q correctly by comparing with the standard form. Verify your integrating factor before applying the general solution formula.

Question 2. Find the general solution for each of the following differential equations.
\( x\frac{dy}{dx} + 2y = x^2 \)
Answer: The given differential equation is \( x\frac{dy}{dx} + 2y = x^2 \).

Dividing the entire equation by x to convert it to standard form:

\( \frac{dy}{dx} + \frac{2}{x}.y = x \)

This is now in the standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2}{x} \) and \( Q = x \).

The integrating factor is:

\( I.F. = e^{\int \frac{2}{x} dx} = e^{2\log x} = e^{\log x^2} = x^2 \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot x^2 = \int x \cdot x^2 \, dx + c \)

\( x^2 y = \int x^3 \, dx + c \)

\( x^2 y = \frac{x^4}{4} + c \)

\( y = \frac{x^2}{4} + \frac{c}{x^2} \)

Exam Tip: Always divide the equation first if the coefficient of \( \frac{dy}{dx} \) is not 1. This step is crucial for finding the correct integrating factor.

Question 3. Find the general solution for each of the following differential equations.
\( 2x\frac{dy}{dx} + y = 6x^3 \)
Answer: The given differential equation is \( 2x\frac{dy}{dx} + y = 6x^3 \).

Dividing the entire equation by 2x to get standard form:

\( \frac{dy}{dx} + \frac{1}{2x}.y = 3x^2 \)

This corresponds to the standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{2x} \) and \( Q = 3x^2 \).

The integrating factor is:

\( I.F. = e^{\int \frac{1}{2x} dx} = e^{\frac{1}{2}\log x} = e^{\log \sqrt{x}} = \sqrt{x} \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot \sqrt{x} = \int 3x^2 \cdot \sqrt{x} \, dx + c \)

\( \sqrt{x} \cdot y = \int 3x^{5/2} \, dx + c \)

\( \sqrt{x} \cdot y = 3 \cdot \frac{x^{7/2}}{7/2} + c \)

\( \sqrt{x} \cdot y = \frac{6x^{7/2}}{7} + c \)

Dividing by \( \sqrt{x} \):

\( y = \frac{6x^3}{7} + \frac{c}{\sqrt{x}} \)

Exam Tip: When the integrating factor involves a root or fractional power, be careful with exponent arithmetic when integrating the product Q \( \cdot \) I.F.

Question 4. Find the general solution for each of the following differential equations.
\( x\frac{dy}{dx} + y = 3x^2 - 2, \quad x > 0 \)
Answer: The given differential equation is \( x\frac{dy}{dx} + y = 3x^2 - 2 \).

Dividing by x to get standard form:

\( \frac{dy}{dx} + \frac{1}{x}.y = \frac{3x^2 - 2}{x} \)

This matches \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x} \) and \( Q = \frac{3x^2 - 2}{x} \).

The integrating factor is:

\( I.F. = e^{\int \frac{1}{x} dx} = e^{\log x} = x \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot x = \int \frac{3x^2 - 2}{x} \cdot x \, dx + c \)

\( xy = \int (3x^2 - 2) \, dx + c \)

\( xy = x^3 - 2x + c \)

Dividing by x:

\( y = x^2 - 2 + \frac{c}{x} \)

Exam Tip: Simplify the expression for Q before multiplying by the integrating factor - this reduces calculation errors and makes the integration step cleaner.

Question 5. Find the general solution for each of the following differential equations.
\( x\frac{dy}{dx} - y = 2x^3 \)
Answer: The given differential equation is \( x\frac{dy}{dx} - y = 2x^3 \).

Dividing by x to obtain standard form:

\( \frac{dy}{dx} - \frac{1}{x}.y = 2x^2 \)

This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{x} \) and \( Q = 2x^2 \).

The integrating factor is:

\( I.F. = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x} \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot \frac{1}{x} = \int 2x^2 \cdot \frac{1}{x} \, dx + c \)

\( \frac{y}{x} = \int 2x \, dx + c \)

\( \frac{y}{x} = x^2 + c \)

Multiplying by x:

\( y = x^3 + cx \)

Exam Tip: When P is negative, ensure the integrating factor calculation includes the negative sign correctly - check by substituting back into the original equation.

Question 6. Find the general solution for each of the following differential equations.
\( x\frac{dy}{dx} - y = x + 1 \)
Answer: The given differential equation is \( x\frac{dy}{dx} - y = x + 1 \).

Dividing by x to get standard form:

\( \frac{dy}{dx} - \frac{1}{x}.y = \frac{x + 1}{x} \)

This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{x} \) and \( Q = \frac{x + 1}{x} \).

The integrating factor is:

\( I.F. = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x} \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot \frac{1}{x} = \int \frac{x + 1}{x} \cdot \frac{1}{x} \, dx + c \)

\( \frac{y}{x} = \int \frac{x + 1}{x^2} \, dx + c \)

\( \frac{y}{x} = \int \left(\frac{1}{x} + \frac{1}{x^2}\right) dx + c \)

\( \frac{y}{x} = \log x - \frac{1}{x} + c \)

Multiplying by x:

\( y = x\log x - 1 + cx \)

Exam Tip: When the right side is a fraction, expand or decompose it before integrating - breaking it into simpler parts makes each integral manageable.

Question 7. Find the general solution for each of the following differential equations.
\( (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \)
Answer: The given differential equation is \( (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2} \).

Dividing by (1 + x^2):

\( \frac{dy}{dx} + \frac{2x}{1 + x^2}.y = \frac{1}{(1 + x^2)^2} \)

This is in standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{1}{(1 + x^2)^2} \).

The integrating factor is:

\( I.F. = e^{\int \frac{2x}{1 + x^2} dx} \)

Let \( f(x) = 1 + x^2 \), so \( f'(x) = 2x \):

\( I.F. = e^{\log(1 + x^2)} = 1 + x^2 \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) \, dx + c \)

\( y \cdot (1 + x^2) = \int \frac{1}{1 + x^2} \, dx + c \)

\( y \cdot (1 + x^2) = \tan^{-1}x + c \)

Therefore:

\( y \cdot (1 + x^2) = \tan^{-1}x + c \)

Exam Tip: Recognize substitution patterns in the integrating factor - if the numerator of P is the derivative of the denominator, the logarithm simplifies to a direct form.

Question 8. Find the general solution for each of the following differential equations.
\( (1 - x^2)\frac{dy}{dx} + xy = x\sqrt{1 - x^2} \)
Answer: The given differential equation is \( (1 - x^2)\frac{dy}{dx} + xy = x\sqrt{1 - x^2} \).

Dividing by (1 - x^2):

\( \frac{dy}{dx} + \frac{x}{1 - x^2}.y = \frac{x}{\sqrt{1 - x^2}} \)

This matches the standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{x}{1 - x^2} \) and \( Q = \frac{x}{\sqrt{1 - x^2}} \).

The integrating factor is:

\( I.F. = e^{\int \frac{x}{1 - x^2} dx} \)

Let \( f(x) = 1 - x^2 \), so \( f'(x) = -2x \):

\( I.F. = e^{-\frac{1}{2}\int \frac{-2x}{1 - x^2} dx} = e^{-\frac{1}{2}\log(1 - x^2)} = e^{\log(1 - x^2)^{-1/2}} = \frac{1}{\sqrt{1 - x^2}} \)

Using the general solution:

\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)

\( y \cdot \frac{1}{\sqrt{1 - x^2}} = \int \frac{x}{\sqrt{1 - x^2}} \cdot \frac{1}{\sqrt{1 - x^2}} \, dx + c \)

\( \frac{y}{\sqrt{1 - x^2}} = \int \frac{x}{1 - x^2} \, dx + c \)

Let \( u = 1 - x^2 \), so \( du = -2x \, dx \):

\( \frac{y}{\sqrt{1 - x^2}} = -\frac{1}{2}\log(1 - x^2) + c \)

Therefore:

\( y = -\frac{\sqrt{1 - x^2}}{2}\log(1 - x^2) + c\sqrt{1 - x^2} \)

Exam Tip: When working with integrating factors involving fractional powers, use substitution techniques to handle the integral efficiently and avoid computational errors.

Question 9. Find the general solution for each of the following differential equations.
\( (1-x^2)\frac{dy}{dx} + xy = ax \)
Answer: The given differential equation is \( (1-x^2)\frac{dy}{dx} + xy = ax \). Dividing by \( (1-x^2) \), we get \( \frac{dy}{dx} + \frac{x}{(1-x^2)} \cdot y = \frac{ax}{(1-x^2)} \). This is in the standard form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{x}{(1-x^2)} \) and \( Q = \frac{ax}{(1-x^2)} \). The integrating factor is found by evaluating \( \int \frac{x}{(1-x^2)} dx \). Letting \( (1-x^2) = f(x) \), we have \( f'(x) = -2x \), so \( \int \frac{x}{(1-x^2)} dx = -\frac{1}{2}\log(1-x^2) \). Thus, \( I.F. = e^{-\frac{1}{2}\log(1-x^2)} = e^{\log(1-x^2)^{-1/2}} = \frac{1}{\sqrt{1-x^2}} \). The general solution is \( y \cdot \frac{1}{\sqrt{1-x^2}} = \int \frac{ax}{(1-x^2)^{3/2}} dx + c \). Using substitution \( (1-x^2) = t \), so \( x\,dx = -\frac{dt}{2} \), we evaluate \( I = \int \frac{ax}{(1-x^2)^{3/2}} dx = -\frac{a}{2} \int t^{-3/2} dt = -\frac{a}{2} \cdot \frac{t^{-1/2}}{-1/2} = \frac{a}{\sqrt{t}} = \frac{a}{\sqrt{1-x^2}} \). Substituting back: \( \frac{y}{\sqrt{1-x^2}} = \frac{a}{\sqrt{1-x^2}} + c \). Multiplying by \( \sqrt{1-x^2} \) gives \( y = a + c\sqrt{1-x^2} \).

Exam Tip: Recognize the standard linear form by dividing through to isolate \( \frac{dy}{dx} \). Verify your integrating factor by checking the logarithmic integral carefully, and always multiply both sides by the I.F. before applying the general solution formula.

Question 10. Find the general solution for each of the following differential equations.
\( (x^2 + 1)\frac{dy}{dx} - 2xy = (x^2 + 1)(x^2 + 2) \)
Answer: The given differential equation is \( (x^2 + 1)\frac{dy}{dx} - 2xy = (x^2 + 1)(x^2 + 2) \). Dividing by \( (1 + x^2) \), we obtain \( \frac{dy}{dx} + \frac{-2x}{(1+x^2)} \cdot y = (x^2 + 2) \). This matches the form \( \frac{dy}{dx} + Py = Q \) with \( P = \frac{-2x}{(1+x^2)} \) and \( Q = (x^2 + 2) \). The integrating factor is \( I.F. = e^{\int P\,dx} = e^{\int \frac{-2x}{(1+x^2)} dx} \). Since \( \int \frac{-2x}{(1+x^2)} dx = -\log(1+x^2) \), we have \( I.F. = e^{-\log(1+x^2)} = \frac{1}{(1+x^2)} \). The general solution is \( y \cdot \frac{1}{(1+x^2)} = \int (x^2 + 2) \cdot \frac{1}{(1+x^2)} dx + c \). Simplifying the integral: \( \int \frac{x^2+2}{1+x^2} dx = \int \left(1 + \frac{1}{1+x^2}\right) dx = x + \tan^{-1}x + c \). Thus, \( \frac{y}{(1+x^2)} = x + \tan^{-1}x + c \), and multiplying by \( (1+x^2) \) gives the general solution \( y = (1 + x^2)(x + \tan^{-1}x + c) \).

Exam Tip: When dividing the equation, identify P and Q precisely. The integral of \( \frac{1}{1+x^2} \) is \( \tan^{-1}x \) - this is a standard formula you must recognize. Always multiply the final relation by the integrating factor's reciprocal to isolate y.

Question 11. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 2y = 6e^x \)
Answer: The given differential equation is \( \frac{dy}{dx} + 2y = 6e^x \), which is already in standard form \( \frac{dy}{dx} + Py = Q \) with \( P = 2 \) and \( Q = 6e^x \). The integrating factor is \( I.F. = e^{\int 2\,dx} = e^{2x} \). The general solution is \( y(e^{2x}) = \int 6e^x \cdot e^{2x} dx + c = 6 \int e^{3x} dx + c \). Evaluating the integral: \( 6 \int e^{3x} dx = 6 \cdot \frac{e^{3x}}{3} = 2e^{3x} + c \). Thus, \( y(e^{2x}) = 2e^{3x} + c \). Dividing both sides by \( e^{2x} \) gives \( y = \frac{2e^{3x}}{e^{2x}} + \frac{c}{e^{2x}} = 2e^x + ce^{-2x} \).

Exam Tip: Recognize when an equation is already in standard form - this saves a step. Use the formula \( \int e^{kx} dx = \frac{e^{kx}}{k} \) confidently, and always simplify exponential fractions by combining exponents: \( e^{a-b} = e^a / e^b \).

Question 12. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 3y = e^{-2x} \)
Answer: The given differential equation is \( \frac{dy}{dx} + 3y = e^{-2x} \), which is in standard form with \( P = 3 \) and \( Q = e^{-2x} \). The integrating factor is \( I.F. = e^{\int 3\,dx} = e^{3x} \). The general solution is \( y(e^{3x}) = \int e^{-2x} \cdot e^{3x} dx + c = \int e^x dx + c = e^x + c \). Thus, \( y(e^{3x}) = e^x + c \). Dividing by \( e^{3x} \) yields \( y = \frac{e^x}{e^{3x}} + \frac{c}{e^{3x}} = e^{-2x} + ce^{-3x} \).

Exam Tip: After multiplying by the integrating factor, the left side \( y \cdot I.F. \) becomes \( \frac{d}{dx}[y \cdot I.F.] \) automatically. Always combine exponential terms in the integral before evaluating, and simplify the final quotient of exponentials carefully.

Question 13. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 8y = 5e^{-3x} \)
Answer: The given differential equation is \( \frac{dy}{dx} + 8y = 5e^{-3x} \), in standard form with \( P = 8 \) and \( Q = 5e^{-3x} \). The integrating factor is \( I.F. = e^{\int 8\,dx} = e^{8x} \). The general solution is \( y(e^{8x}) = \int 5e^{-3x} \cdot e^{8x} dx + c = 5 \int e^{5x} dx + c = 5 \cdot \frac{e^{5x}}{5} + c = e^{5x} + c \). Thus, \( y(e^{8x}) = e^{5x} + c \). Dividing by \( e^{8x} \) gives \( y = \frac{e^{5x}}{e^{8x}} + \frac{c}{e^{8x}} = e^{-3x} + ce^{-8x} \).

Exam Tip: Check your exponent arithmetic: when you combine \( e^{-3x} \) and \( e^{8x} \), the result is \( e^{5x} \). Simplify the final fraction carefully by subtracting exponents in the correct order to avoid sign errors.

Question 14. Find the general solution for each of the following differential equations.
\( x\frac{dy}{dx} - y = (x - 1)e^x, \quad x > 0 \)
Answer: The given differential equation is \( x\frac{dy}{dx} - y = (x - 1)e^x \). Dividing by x, we get \( \frac{dy}{dx} - \frac{1}{x} \cdot y = \frac{(x-1)e^x}{x} \), which is in standard form with \( P = -\frac{1}{x} \) and \( Q = \frac{(x-1)e^x}{x} \). The integrating factor is \( I.F. = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x} \). The general solution is \( y \cdot \frac{1}{x} = \int \frac{(x-1)e^x}{x} \cdot \frac{1}{x} dx + c = \int \frac{(x-1)e^x}{x^2} dx + c \). This integral is evaluated by rewriting \( \frac{x-1}{x^2} = \frac{1}{x} - \frac{1}{x^2} \), so \( \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx \). Using the formula \( \int e^x(f(x) + f'(x)) dx = e^x f(x) \), with \( f(x) = \frac{1}{x} \) and \( f'(x) = -\frac{1}{x^2} \), we get \( I = e^x \cdot \frac{1}{x} \). Thus, \( \frac{y}{x} = e^x \cdot \frac{1}{x} + c \). Multiplying by x gives the general solution \( y = e^x + cx \).

Exam Tip: Recognize integrals of the form \( \int e^x(f(x) + f'(x)) dx = e^x f(x) \) - this is a key technique. Always verify that your function f and its derivative match the terms inside the integrand before applying the formula.

Question 15. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} - y \tan x = e^x \sec x \)
Answer: The given differential equation is \( \frac{dy}{dx} - y \tan x = e^x \sec x \), in standard form with \( P = -\tan x \) and \( Q = e^x \sec x \). The integrating factor is \( I.F. = e^{\int -\tan x\,dx} = e^{-(-\log(\sec x))} = e^{\log(\sec x)} = \sec x \). The general solution is \( y \cdot \sec x = \int e^x \sec x \cdot \sec x\, dx + c = \int e^x \sec^2 x\, dx + c \). To evaluate this integral, we use the fact that \( \sec^2 x = 1 + \tan x \cdot \frac{d}{dx}[\sec x] \), or recognize that \( \int e^x (\tan x + \sec x \tan x) dx = e^x \sec x \) (using the product rule derivative). Thus, \( \int e^x \sec^2 x\, dx = e^x \sec x \). Therefore, \( y \cdot \sec x = e^x \sec x + c \). Dividing by \( \sec x \) gives the general solution \( y = e^x + c \cos x \).

Exam Tip: The integral \( \int \tan x\, dx = -\log(\sec x) + C \) or equivalently \( \log(\cos x) + C \) is essential - memorize it. Recognize when an exponential-trigonometric product integral matches the form \( \int e^x(f(x) + f'(x)) dx \), and apply the formula systematically.

Question 16. Find the general solution for each of the following differential equations.
\( (x \log x) \frac{dy}{dx} + y = 2 \log x \)
Answer: The given differential equation is \( (x \log x) \frac{dy}{dx} + y = 2 \log x \).

Divide the above equation by (x.log x):
\( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x} \) .....eq(1)

This equation takes the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x \log x} \) and \( Q = \frac{2}{x} \).

The integrating factor is calculated as:
\( I.F. = e^{\int P \, dx} = e^{\int \frac{1}{x \log x} dx} = e^{\int \frac{1/x}{\log x} dx} \)

Let \( f(x) = \log x \), so \( f'(x) = \frac{1}{x} \):
\( I.F. = e^{\log(\log x)} \) (using \( \int \frac{f'(x)}{f(x)} dx = \log(f(x)) \)) = \( \log x \) (using \( a^{\log_a b} = b \))

The general solution is given by:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot (\log x) = \int \frac{2}{x} \cdot (\log x) \, dx + c \)
\( y \cdot (\log x) = 2 \int \frac{1}{x} \log x \, dx + c \) .....eq(2)

To evaluate \( I = \int \frac{1}{x} \log x \, dx \), use integration by parts with \( u = \log x \) and \( v = \frac{1}{x} \):
\( I = \log x \int \frac{1}{x} dx - \int \frac{d}{dx}(\log x) \cdot \int \frac{1}{x} dx \, dx \)
\( I = \log x \cdot \log x - \int \frac{1}{x} \cdot \log x \, dx \)
\( I = (\log x)^2 - I \)
\( 2I = (\log x)^2 \)
\( I = \frac{1}{2} (\log x)^2 \)

Substituting I in eq(2):
\( y \cdot (\log x) = 2 \cdot \frac{1}{2} (\log x)^2 + c \)
\( y \cdot (\log x) = (\log x)^2 + c \)

The general solution is \( y \cdot (\log x) = (\log x)^2 + c \).
In simple words: Start by rearranging the equation into standard form by dividing through. The integrating factor helps you simplify and combine terms. Use integration by parts to work through the integral, and you will reach the final answer.

Exam Tip: Always identify the equation form first, calculate the integrating factor carefully using logarithm properties, and use integration by parts when products of logarithms appear.

Question 17. Find the general solution for each of the following differential equations.
\( x \frac{dy}{dx} + y = x \log x \)
Answer: The given differential equation is \( x \frac{dy}{dx} + y = x \log x \).

Divide the above equation by x:
\( \frac{dy}{dx} + \frac{1}{x} y = \log x \) .....eq(1)

This equation is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x} \) and \( Q = \log x \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int \frac{1}{x} dx} = e^{\log x} \) (using \( \int \frac{1}{x} dx = \log x \)) = \( x \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot x = \int (x \log x) dx + c \) .....eq(2)

To evaluate \( I = \int (x \log x) dx \), use integration by parts with \( u = \log x \) and \( v = x \):
\( I = \log x \int x \, dx - \int \frac{d}{dx}(\log x) \cdot \int x \, dx \, dx \)
\( I = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx \)
\( I = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \)
\( I = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} \)
\( I = \frac{x^2}{2} \log x - \frac{x^2}{4} \)

Substituting I in eq(2):
\( xy = \frac{x^2}{2} \log x - \frac{x^2}{4} + c \)

Multiply the above equation by 4:
\( 4xy = 2x^2 \log x - x^2 + 4c \)

The general solution is \( 4xy = 2x^2 \log x - x^2 + 4c \).
In simple words: Convert to standard form, find the integrating factor, apply the general solution formula, and use integration by parts to evaluate the product integral.

Exam Tip: Watch for products involving logarithms and be careful with the integration by parts formula; multiply the final result by a constant if needed to simplify.

Question 18. Find the general solution for each of the following differential equations.
\( x \frac{dy}{dx} + 2y = x^2 \log x \)
Answer: The given differential equation is \( x \frac{dy}{dx} + 2y = x^2 \log x \).

Divide the above equation by x:
\( \frac{dy}{dx} + \frac{2}{x} y = x \log x \) .....eq(1)

This equation takes the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2}{x} \) and \( Q = x \log x \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int \frac{2}{x} dx} = e^{2 \int \frac{1}{x} dx} = e^{2 \log x} \) (using \( \int \frac{1}{x} dx = \log x \)) = \( e^{\log x^2} \) (using \( a \log b = \log b^a \)) = \( x^2 \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot (x^2) = \int (x^2 \cdot x \log x) dx + c \)
\( y \cdot (x^2) = \int (x^3 \log x) dx + c \) .....eq(2)

To evaluate \( I = \int (x^3 \log x) dx \), use integration by parts with \( u = \log x \) and \( v = x^3 \):
\( I = \log x \int x^3 \, dx - \int \frac{d}{dx}(\log x) \cdot \int x^3 \, dx \, dx \)
\( I = \log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx \)
\( I = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx \)
\( I = \frac{x^4}{4} \log x - \frac{1}{4} \cdot \frac{x^4}{4} \)
\( I = \frac{x^4}{4} \log x - \frac{x^4}{16} \)

Substituting I in eq(2):
\( x^2 y = \frac{x^4}{4} \log x - \frac{x^4}{16} + c \)

Divide the above equation by \( x^2 \):
\( y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{c}{x^2} \)
\( y = \frac{x^2}{16} (4 \log x - 1) + \frac{c}{x^2} \)

The general solution is \( y = \frac{x^2}{16} (4 \log x - 1) + \frac{c}{x^2} \).
In simple words: Transform the equation into standard form, calculate the integrating factor which becomes \( x^2 \), and then solve using the general solution method along with integration by parts.

Exam Tip: The integrating factor involves powers of x; careful algebraic manipulation and correct application of the integration by parts formula are essential for success.

Question 19. Find the general solution for each of the following differential equations.
\( (1 + x) \frac{dy}{dx} - y = e^{3x} (1 + x)^2 \)
Answer: The given differential equation is \( (1 + x) \frac{dy}{dx} - y = e^{3x} (1 + x)^2 \).

Divide the above equation by (1 + x):
\( \frac{dy}{dx} - \frac{1}{(1+x)} y = e^{3x} (1 + x) \) .....eq(1)

This equation is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{(1+x)} \) and \( Q = e^{3x} (1 + x) \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int -\frac{1}{(1+x)} dx} = e^{- \int \frac{1}{(1+x)} dx} \)
\( = e^{- \log(1+x)} \) (using \( \int \frac{1}{px+q} dx = \frac{1}{p} \log(px + q) \)) = \( e^{\log(1+x)^{-1}} \) (using \( a \log b = \log b^a \)) = \( \frac{1}{(1+x)} \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot \left( \frac{1}{1+x} \right) = \int e^{3x} (1 + x) \cdot \left( \frac{1}{1+x} \right) dx + c \)
\( y \cdot \left( \frac{1}{1+x} \right) = \int e^{3x} \, dx + c \)
\( y \cdot \left( \frac{1}{1+x} \right) = \frac{1}{3} e^{3x} + c \) (using \( \int e^{kx} dx = \frac{1}{k} e^{kx} \))

Multiply the above equation by (1 + x):
\( y = \frac{1}{3} (1 + x) e^{3x} + c(1 + x) \)

The general solution is \( y = \frac{1}{3} (1 + x) e^{3x} + c(1 + x) \).
In simple words: Rearrange to standard form, find the integrating factor which simplifies to \( \frac{1}{1+x} \), and solve using the general solution formula with straightforward exponential integration.

Exam Tip: When dividing by a polynomial factor, the integrating factor often involves logarithms; carefully apply logarithm properties to simplify the exponential.

Question 20. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + \frac{4x}{(x^2 + 1)} y + \frac{1}{(1 + x^2)^2} = 0 \)
Answer: The given differential equation is \( \frac{dy}{dx} + \frac{4x}{(x^2 + 1)} y + \frac{1}{(1 + x^2)^2} = 0 \).

Rearranging:
\( \frac{dy}{dx} + \frac{4x}{(x^2 + 1)} y = \frac{-1}{(1 + x^2)^2} \) .....eq(1)

This equation is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{4x}{(x^2+1)} \) and \( Q = \frac{-1}{(1+x^2)^2} \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int \frac{4x}{(x^2+1)} dx} = e^{2 \int \frac{2x}{(x^2+1)} dx} \)

Let \( f(x) = (x^2 + 1) \) and \( f'(x) = 2x \):
\( I.F. = e^{2 \log(x^2+1)} \) (using \( \int \frac{f'(x)}{f(x)} dx = \log(f(x)) \)) = \( e^{\log(x^2+1)^2} \) (using \( a \log b = \log b^a \)) = \( (1 + x^2)^2 \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot (1 + x^2)^2 = \int \frac{-1}{(1 + x^2)^2} \cdot (1 + x^2)^2 \, dx + c \)
\( y \cdot (1 + x^2)^2 = \int -1 \, dx + c \)
\( y \cdot (1 + x^2)^2 = -x + c \) (using \( \int 1 \, dx = x \))

Divide the above equation by \( (1+x^2)^2 \):
\( y = \frac{-x}{(1 + x^2)^2} + \frac{c}{(1 + x^2)^2} \)

The general solution is \( y = \frac{-x}{(1 + x^2)^2} + \frac{c}{(1 + x^2)^2} \).
In simple words: Convert to standard form, calculate the integrating factor using logarithm and exponential identities, and the integral simplifies to a basic linear term.

Exam Tip: Notice that the integrating factor exactly cancels one of the terms in the integral; always check if such cancellation occurs before computing.

Question 21. Find the general solution for each of the following differential equations.
\( (y + 3x^2) \frac{dx}{dy} = x \)
Answer: The given differential equation is \( (y + 3x^2) \frac{dx}{dy} = x \).

Rearranging:
\( \frac{dy}{dx} = \frac{y + 3x^2}{x} \)
\( \frac{dy}{dx} = \frac{y}{x} + 3x \)
\( \frac{dy}{dx} - \frac{y}{x} = 3x \) .....eq(1)

This equation is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{x} \) and \( Q = 3x \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int -\frac{1}{x} dx} = e^{- \log x} \) (using \( \int \frac{1}{x} dx = \log x \)) = \( e^{\log(x)^{-1}} \) (using \( a \log b = \log b^a \)) = \( \frac{1}{x} \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot \left( \frac{1}{x} \right) = \int 3x \cdot \left( \frac{1}{x} \right) dx + c \)
\( \frac{y}{x} = \int 3 \, dx + c \)
\( \frac{y}{x} = 3x + c \) (using \( \int 1 \, dx = x \))

Multiply the above equation by x:
\( y = 3x^2 + cx \)

The general solution is \( y = 3x^2 + cx \).
In simple words: Reverse the roles of x and y by rearranging the equation, convert to standard linear form, find the integrating factor, and integrate to get the result.

Exam Tip: When the differential involves \( \frac{dx}{dy} \), rearrange to get \( \frac{dy}{dx} \) before applying the standard method.

Question 22. Find the general solution for each of the following differential equations.
\( x dy - (y + 2x^2) dx = 0 \)
Answer: The given differential equation is \( x dy - (y + 2x^2) dx = 0 \).

Rearranging:
\( x \frac{dy}{dx} - (y + 2x^2) = 0 \)
\( x \frac{dy}{dx} = y + 2x^2 \)
\( \frac{dy}{dx} = \frac{y + 2x^2}{x} \)
\( \frac{dy}{dx} = \frac{y}{x} + 2x \)
\( \frac{dy}{dx} - \frac{y}{x} = 2x \) .....eq(1)

This equation is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{x} \) and \( Q = 2x \).

The integrating factor is:
\( I.F. = e^{\int P \, dx} = e^{\int -\frac{1}{x} dx} = e^{- \log x} \) (using \( \int \frac{1}{x} dx = \log x \)) = \( e^{\log(x)^{-1}} \) (using \( a \log b = \log b^a \)) = \( \frac{1}{x} \) (using \( a^{\log_a b} = b \))

The general solution is:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c \)
\( y \cdot \left( \frac{1}{x} \right) = \int 2x \cdot \left( \frac{1}{x} \right) dx + c \)
\( \frac{y}{x} = \int 2 \, dx + c \)
\( \frac{y}{x} = 2x + c \) (using \( \int 1 \, dx = x \))

Multiply the above equation by x:
\( y = 2x^2 + cx \)

The general solution is \( y = 2x^2 + cx \).
In simple words: Expand the differential form, rearrange to standard linear form, compute the integrating factor, solve using the general solution formula, and simplify.

Exam Tip: Always expand products in differentials carefully and separate variables systematically to avoid algebraic errors.

Question 22. Find the general solution for each of the following differential equations.
\( xdy - (y + 2x^2)dx = 0 \)
Answer: The given differential equation can be rewritten as \( xdy = (y + 2x^2)dx \), which simplifies to \( \frac{dy}{dx} = \frac{y + 2x^2}{x} \) or \( \frac{dy}{dx} = \frac{y}{x} + 2x \).

Rearranging: \( \frac{dy}{dx} - \frac{y}{x} = 2x \) ........eq(1)

This is in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = -\frac{1}{x} \) and \( Q = 2x \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(1/x)} = \frac{1}{x} \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.\left(\frac{1}{x}\right) = \int 2x.\left(\frac{1}{x}\right)dx + c \)

\( \frac{y}{x} = \int 2dx + c \)

\( \frac{y}{x} = 2x + c \)

Multiplying through by \( x \):

\( y = 2x^2 + cx \)

Therefore, the general solution is \( y = 2x^2 + cx \)

Exam Tip: Always rewrite the equation in standard linear form \( \frac{dy}{dx} + Py = Q \) before finding the integrating factor. Verify your solution by differentiating and substituting back into the original equation.

Question 23. Find the general solution for each of the following differential equations.
\( xdy + (y - x^3)dx = 0 \)
Answer: The given differential equation can be rewritten as \( xdy = -(y - x^3)dx = (x^3 - y)dx \), which simplifies to \( \frac{dy}{dx} = \frac{x^3 - y}{x} \) or \( \frac{dy}{dx} = x^2 - \frac{y}{x} \).

Rearranging: \( \frac{dy}{dx} + \frac{y}{x} = x^2 \) ........eq(1)

This is in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = \frac{1}{x} \) and \( Q = x^2 \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(x) = \int x^2.(x)dx + c \)

\( xy = \int x^3dx + c \)

\( xy = \frac{x^4}{4} + c \)

Dividing through by \( x \):

\( y = \frac{x^3}{4} + \frac{c}{x} \)

Therefore, the general solution is \( y = \frac{x^3}{4} + \frac{c}{x} \)

Exam Tip: After finding the integrating factor, carefully integrate the right-hand side. Remember to include the constant of integration and simplify the final answer by dividing by the integrating factor if needed.

Question 24. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 2y = \sin x \)
Answer: The given differential equation is already in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = 2 \) and \( Q = \sin x \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int 2 dx} = e^{2x} \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(e^{2x}) = \int \sin x.(e^{2x})dx + c \) ........eq(2)

Let \( I = \int \sin x.(e^{2x})dx \). Using integration by parts twice with \( u = \sin x \) and \( v = e^{2x} \):

\( I = \sin x.\frac{e^{2x}}{2} - \int \cos x.\frac{e^{2x}}{2}dx = \sin x.\frac{e^{2x}}{2} - \frac{1}{2}\int (\cos x.e^{2x})dx \)

Now using integration by parts again with \( u = \cos x \) and \( v = e^{2x} \):

\( I = \sin x.\frac{e^{2x}}{2} - \frac{1}{2}\left[\cos x.\frac{e^{2x}}{2} - \int (-\sin x).\frac{e^{2x}}{2}dx\right] \)

\( I = \sin x.\frac{e^{2x}}{2} - \frac{1}{2}\left[\cos x.\frac{e^{2x}}{2} + \frac{1}{2}\int (\sin x.e^{2x})dx\right] \)

\( I = \sin x.\frac{e^{2x}}{2} - \frac{1}{2}\cos x.\frac{e^{2x}}{2} - \frac{1}{4}I \)

\( I + \frac{1}{4}I = \sin x.\frac{e^{2x}}{2} - \cos x.\frac{e^{2x}}{4} \)

\( \frac{5}{4}I = \frac{e^{2x}}{4}(2\sin x - \cos x) \)

\( I = \frac{e^{2x}}{5}(2\sin x - \cos x) \)

Substituting into eq(2):

\( y.(e^{2x}) = \frac{e^{2x}}{5}(2\sin x - \cos x) + c \)

Dividing by \( e^{2x} \):

\( y = \frac{1}{5}(2\sin x - \cos x) + ce^{-2x} \)

Therefore, the general solution is \( y = \frac{1}{5}(2\sin x - \cos x) + ce^{-2x} \)

Exam Tip: Integration by parts for products of exponential and trigonometric functions requires applying the formula twice. Collect the original integral term on one side and solve algebraically to avoid repeating the calculation indefinitely.

Question 25. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + y = \cos x - \sin x \)
Answer: The given differential equation is already in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = 1 \) and \( Q = \cos x - \sin x \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int 1 dx} = e^x \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(e^x) = \int (\cos x - \sin x).(e^x)dx + c \)

Using the integration formula \( \int e^x(f(x) + f'(x))dx = e^x.f(x) \) with \( f(x) = \cos x \) (so \( f'(x) = -\sin x \)):

\( y.(e^x) = \int e^x.(\cos x + (-\sin x))dx + c = e^x.\cos x + c \)

Dividing by \( e^x \):

\( y = \cos x + ce^{-x} \)

Therefore, the general solution is \( y = \cos x + ce^{-x} \)

Exam Tip: Recognize the special formula \( \int e^x(f(x) + f'(x))dx = e^x.f(x) \) - when the integrand has this form, the integral simplifies directly. This saves time and reduces calculation errors.

Question 26. Find the general solution for each of the following differential equations.
\( \sec x \frac{dy}{dx} - y = \sin x \)
Answer: The given differential equation can be rewritten by dividing through by \( \sec x \):

\( \frac{dy}{dx} - \frac{1}{\sec x}y = \frac{\sin x}{\sec x} \)

\( \frac{dy}{dx} - \cos x.y = \sin x.\cos x \) ........eq(1)

This is in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = -\cos x \) and \( Q = \sin x.\cos x \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int -\cos x dx} = e^{-\sin x} \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(e^{-\sin x}) = \int (\sin x.\cos x).(e^{-\sin x})dx + c \) ........eq(2)

Let \( I = \int (\sin x.\cos x).(e^{-\sin x})dx \). Using substitution \( t = \sin x \) so \( \cos x.dx = dt \):

\( I = \int t.e^{-t}dt \)

Using integration by parts with \( u = t \) and \( dv = e^{-t}dt \):

\( I = t.\int e^{-t}dt - \int \frac{d}{dt}(t).\left(\int e^{-t}dt\right)dt \)

\( I = -t.e^{-t} - \int 1.(-e^{-t})dt = -t.e^{-t} + \int e^{-t}dt \)

\( I = -t.e^{-t} - e^{-t} = -e^{-t}(t + 1) = -\sin x.e^{-\sin x} - e^{-\sin x} \)

Substituting into eq(2):

\( y.(e^{-\sin x}) = -\sin x.e^{-\sin x} - e^{-\sin x} + c \)

\( y.(e^{-\sin x}) = -e^{-\sin x}(\sin x + 1) + c \)

Dividing by \( e^{-\sin x} \):

\( y = -(\sin x + 1) + ce^{\sin x} = ce^{\sin x} - (\sin x + 1) \)

Therefore, the general solution is \( y = ce^{\sin x} - (\sin x + 1) \)

Exam Tip: Always check if a substitution simplifies the integral after multiplying by the integrating factor. Substitutions like \( t = \sin x \) or \( t = \cos x \) often convert trigonometric integrals into standard polynomial forms.

Question 27. Find the general solution for each of the following differential equations.
\( (1 + x^2)\frac{dy}{dx} + 2xy = \cot x \)
Answer: The given differential equation can be rewritten by dividing through by \( (1 + x^2) \):

\( \frac{dy}{dx} + \frac{2x}{(1 + x^2)}y = \frac{\cot x}{(1 + x^2)} \) ........eq(1)

This is in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = \frac{2x}{(1 + x^2)} \) and \( Q = \frac{\cot x}{(1 + x^2)} \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int \frac{2x}{(1 + x^2)} dx} \)

Let \( f(x) = (1 + x^2) \), then \( f'(x) = 2x \):

\( I.F. = e^{\log(1 + x^2)} = (1 + x^2) \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(1 + x^2) = \int \frac{\cot x}{(1 + x^2)}.(1 + x^2)dx + c \)

\( y.(1 + x^2) = \int \cot x dx + c \)

\( y.(1 + x^2) = \log|\sin x| + c \)

Therefore, the general solution is \( y.(1 + x^2) = \log|\sin x| + c \)

Exam Tip: Recognize the logarithmic derivative pattern: when \( P = \frac{f'(x)}{f(x)} \), the integrating factor becomes \( I.F. = e^{\log f(x)} = f(x) \). This significantly simplifies the subsequent integration step.

Question 28. Find the general solution for each of the following differential equations.
\( (\sin x)\frac{dy}{dx} + (\cos x)y = \cos x.\sin^2 x \)
Answer: The given differential equation can be rewritten by dividing through by \( \sin x \):

\( \frac{dy}{dx} + \frac{\cos x}{\sin x}y = \cos x.\sin x \)

\( \frac{dy}{dx} + (\cot x)y = \cos x.\sin x \) ........eq(1)

This is in the standard form \( \frac{dy}{dx} + Py = Q \) where \( P = \cot x \) and \( Q = \sin x.\cos x \).

The integrating factor is:

\( I.F. = e^{\int P \, dx} = e^{\int \cot x dx} = e^{\log(\sin x)} = \sin x \)

The general solution is given by:

\( y.(I.F.) = \int Q.(I.F.)dx + c \)

\( y.(\sin x) = \int (\sin x.\cos x).(\sin x)dx + c \)

\( y.(\sin x) = \int (\sin^2 x.\cos x)dx + c \) ........eq(2)

Let \( I = \int (\sin^2 x.\cos x)dx \). Using substitution \( t = \sin x \) so \( \cos x.dx = dt \):

\( I = \int t^2 dt = \frac{t^3}{3} = \frac{\sin^3 x}{3} \)

Substituting back into eq(2):

\( y.(\sin x) = \frac{\sin^3 x}{3} + c \)

Dividing by \( \sin x \):

\( y = \frac{\sin^2 x}{3} + \frac{c}{\sin x} \)

Therefore, the general solution is \( y = \frac{\sin^2 x}{3} + \frac{c}{\sin x} \)

Exam Tip: After multiplying by the integrating factor, look for substitution opportunities in the integral. A power of trigonometric function times its derivative is a classic setup for direct substitution and quick integration.

Question 29. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 2y \cot x = 3x^2 \cos ec^2 x \)
Answer: The given differential equation is \( \frac{dy}{dx} + 2y(\cot x) = 3x^2 \cosec^2 x \). This equation has the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2 \cot x \) and \( Q = 3x^2 \cosec^2 x \). The integrating factor is \( e^{\int P \, dx} = e^{\int 2 \cot x \, dx} = e^{2 \log(\sin x)} = e^{\log(\sin x)^2} = \sin^2 x \). The general solution is given by \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) \, dx + c \). Thus, \( y \cdot (\sin^2 x) = \int (3x^2 \cosec^2 x) \cdot (\sin^2 x) \, dx + c = \int 3x^2 \, dx + c = x^3 + c \). Therefore, the general solution is \( y(\sin^2 x) = x^3 + c \).

Exam Tip: Always identify P and Q correctly, compute the integrating factor systematically, and simplify the resulting integral before solving for y.

Question 30. Find the general solution for each of the following differential equations.
\( x \frac{dy}{dx} - y = 2x^2 \sec x \)
Answer: The given differential equation is \( x \frac{dy}{dx} - y = 2x^2 \sec x \). Dividing by x: \( \frac{dy}{dx} - \frac{1}{x} y = 2x \sec x \). This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\frac{1}{x} \) and \( Q = 2x \sec x \). The integrating factor is \( e^{\int P \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x} \). The general solution is \( y \cdot \left(\frac{1}{x}\right) = \int (2x \sec x) \cdot \left(\frac{1}{x}\right) \, dx + c = \int 2 \sec x \, dx + c = 2 \log|\sec x + \tan x| + c \). Multiplying by x: \( y = 2x \log|\sec x + \tan x| + cx \). Therefore, the general solution is \( y = 2x \log|\sec x + \tan x| + cx \).

Exam Tip: When the coefficient of \( \frac{dy}{dx} \) is not 1, first divide to standardize the equation before finding the integrating factor.

Question 31. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} = y \tan x - 2 \sin x \)
Answer: The given differential equation is \( \frac{dy}{dx} = y \tan x - 2 \sin x \). Rearranging: \( \frac{dy}{dx} - y \tan x = -2 \sin x \). This has the form \( \frac{dy}{dx} + Py = Q \), where \( P = -\tan x \) and \( Q = -2 \sin x \). The integrating factor is \( e^{\int P \, dx} = e^{\int -\tan x \, dx} = e^{-\log|\sec x|} = e^{\log|\cos x|} = \cos x \). The general solution is \( y \cdot (\cos x) = \int (-2 \sin x) \cdot (\cos x) \, dx + c = -\int 2 \sin x \cos x \, dx + c = -\int \sin 2x \, dx + c = \frac{\cos 2x}{2} + c \). Multiplying by 2: \( 2y(\cos x) = \cos 2x + 2c \), which gives \( 2y(\cos x) = \cos 2x + C \) where \( C = 2c \). Therefore, the general solution is \( 2y(\cos x) = \cos 2x + C \).

Exam Tip: Use double angle identities to simplify products like \( \sin x \cos x \), and always simplify the final answer by multiplying through if needed.

Question 32. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} = y \cot x = \sin 2x \)
Answer: The given differential equation is \( \frac{dy}{dx} + y \cot x = \sin 2x \). This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \cot x \) and \( Q = \sin 2x \). The integrating factor is \( e^{\int P \, dx} = e^{\int \cot x \, dx} = e^{\log|\sin x|} = \sin x \). The general solution is \( y(\sin x) = \int (\sin 2x)(\sin x) \, dx + c \). Let \( I = \int (\sin 2x)(\sin x) \, dx \). Using integration by parts with \( u = \sin 2x \) and \( v = \sin x \): \( I = \sin 2x \int \sin x \, dx - \int \frac{d}{dx}(\sin 2x) \int \sin x \, dx \, dx = -\sin 2x \cos x - \int (2 \cos 2x)(-\cos x) \, dx \) \( = -\sin 2x \cos x + 2 \int (\cos 2x)(\cos x) \, dx \). For the second integral, let \( u = \cos 2x \) and \( v = \cos x \): \( \int (\cos 2x)(\cos x) \, dx = \cos 2x \sin x - \int (-2 \sin 2x)(\sin x) \, dx = \cos 2x \sin x + 2 \int (\sin 2x)(\sin x) \, dx \). So \( I = -\sin 2x \cos x + 2 \cos 2x \sin x + 4I \), which gives \( -3I = -\sin 2x \cos x + 2 \cos 2x \sin x \). Thus \( I = \frac{2 \sin 2x \cos x - 2 \cos 2x \sin x}{3} = \frac{2 \sin x}{3} \). Substituting: \( y(\sin x) = \frac{2 \sin^3 x}{3} + c \). Therefore, the general solution is \( y(\sin x) = \frac{2 \sin^3 x}{3} + c \).

Exam Tip: When products of trigonometric functions appear in integrals, integration by parts combined with algebraic manipulation can often simplify the problem.

Question 33. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + 2y \tan x = \sin x \)
Answer: The given differential equation is \( \frac{dy}{dx} + 2y \tan x = \sin x \). This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = 2 \tan x \) and \( Q = \sin x \). The integrating factor is \( e^{\int P \, dx} = e^{\int 2 \tan x \, dx} = e^{2 \log|\sec x|} = e^{\log|\sec x|^2} = \sec^2 x = \frac{1}{\cos^2 x} \). The general solution is \( y \cdot \left(\frac{1}{\cos^2 x}\right) = \int (\sin x) \cdot \left(\frac{1}{\cos^2 x}\right) \, dx + c \). Let \( I = \int (\sin x) \cdot \left(\frac{1}{\cos^2 x}\right) \, dx \). Substitute \( \cos x = t \), so \( -\sin x \, dx = dt \): \( I = \int \frac{-1}{t^2} \, dt = \frac{1}{t} = \frac{1}{\cos x} \). Thus \( y \cdot \left(\frac{1}{\cos^2 x}\right) = \frac{1}{\cos x} + c \). Multiplying by \( \cos^2 x \): \( y = \cos x + c \cos^2 x \). Therefore, the general solution is \( y = \cos x + c(\cos^2 x) \).

Exam Tip: Substitution is a powerful technique for simplifying integrals; always check if a u-substitution can reduce the complexity of the integrand.

Question 34. Find the general solution for each of the following differential equations.
\( \frac{dy}{dx} + y \cot x = x^2 \cot x + 2x \)
Answer: The given differential equation is \( \frac{dy}{dx} + y \cot x = x^2 \cot x + 2x \). This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \cot x \) and \( Q = x^2 \cot x + 2x \). The integrating factor is \( e^{\int P \, dx} = e^{\int \cot x \, dx} = e^{\log|\sin x|} = \sin x \). The general solution is \( y(\sin x) = \int (x^2 \cot x + 2x)(\sin x) \, dx + c = \int (x^2 \cos x + 2x \sin x) \, dx + c \). Let \( I = \int x^2 \cos x \, dx \). Using integration by parts with \( u = x^2 \) and \( v = \cos x \): \( I = x^2 \sin x - \int 2x \sin x \, dx \). For \( \int 2x \sin x \, dx \), use integration by parts again with \( u = 2x \) and \( v = \sin x \): \( \int 2x \sin x \, dx = -2x \cos x + \int 2 \cos x \, dx = -2x \cos x + 2 \sin x \). Thus \( I = x^2 \sin x - (-2x \cos x + 2 \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x \). Also, \( \int 2x \sin x \, dx = -2x \cos x + 2 \sin x \). So the total integral is \( (x^2 \sin x + 2x \cos x - 2 \sin x) + (-2x \cos x + 2 \sin x) = x^2 \sin x \). Thus \( y(\sin x) = x^2 \sin x + c \). Dividing by \( \sin x \): \( y = x^2 + c \cosec x \). Therefore, the general solution is \( y = x^2 + c(\cos ec x) \).

Exam Tip: When the right side contains a product involving the coefficient P, careful expansion and successive integration by parts are required; organize your work clearly.

Question 35. Find a particular solution satisfying the given condition for each of the following differential equations.
\( x \frac{dy}{dx} + y = x^3 \), given that \( y = 1 \) when \( x = 2 \)
Answer: The given differential equation is \( x \frac{dy}{dx} + y = x^3 \). Dividing by x: \( \frac{dy}{dx} + \frac{1}{x} y = x^2 \). This is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x} \) and \( Q = x^2 \). The integrating factor is \( e^{\int P \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \). The general solution is \( y \cdot x = \int x^2 \cdot x \, dx + c = \int x^3 \, dx + c = \frac{x^4}{4} + c \). Thus \( y = \frac{x^3}{4} + \frac{c}{x} \). Using the condition \( y = 1 \) when \( x = 2 \): \( 1 = \frac{8}{4} + \frac{c}{2} = 2 + \frac{c}{2} \), which gives \( \frac{c}{2} = -1 \), so \( c = -2 \). Therefore, the particular solution is \( y = \frac{x^3}{4} - \frac{2}{x} \).

Exam Tip: Always substitute the initial conditions into the general solution to find the constant c, and verify your answer by checking that the initial condition is satisfied.

Question 36. Find a particular solution satisfying the given condition for each of the following differential equations.
\( \frac{dy}{dx} + y \cot x = 4x \cos \text{ec } x \), given that \( y = 0 \) when \( x = \frac{\pi}{2} \).
Answer: The differential equation provided is \( \frac{dy}{dx} + y \cot x = 4x \cos \text{ec } x \).

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = \cot x \) and \( Q = 4x \cos \text{ec } x \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int \cot x \, dx} = e^{\log|\sin x|} = \sin x \)

The general solution is given by:

\( y.(\sin x) = \int (4x \cos \text{ec } x).(\sin x)dx + c \)

\( y.(\sin x) = 4\int \left(x \cdot \frac{1}{\sin x}\right).(\sin x)dx + c \)

\( y.(\sin x) = 4\int x \, dx + c \)

\( y.(\sin x) = 4 \cdot \frac{x^2}{2} + c = 2x^2 + c \)

The general equation becomes \( y.(\sin x) = 2x^2 + c \).

To find the particular solution, substitute \( y = 0 \) and \( x = \frac{\pi}{2} \):

\( 0 = 2 \left(\frac{\pi}{2}\right)^2 + c \)

\( c = -\frac{\pi^2}{2} \)

Therefore, the particular solution is \( y.(\sin x) = 2x^2 - \frac{\pi^2}{2} \).
In simple words: Solve the linear differential equation by finding an integrating factor, then integrate both sides. Apply the initial condition to determine the constant of integration.

Exam Tip: Always verify that your integrating factor is computed correctly by taking its derivative - it should match the coefficient of \( y \) when multiplied by the equation.

Question 37. Find a particular solution satisfying the given condition for each of the following differential equations.
\( \frac{dy}{dx} + 2xy = x \), given that \( y = 0 \) when \( x = 0 \).
Answer: The differential equation provided is \( \frac{dy}{dx} + 2xy = x \).

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = 2x \) and \( Q = x \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int 2x \, dx} = e^{x^2} \)

The general solution is given by:

\( y.(e^{x^2}) = \int x.(e^{x^2})dx + c \)

Let \( I = \int x.(e^{x^2})dx \). Using substitution \( x^2 = t \), we get \( 2x \, dx = dt \):

\( I = \int \frac{1}{2}e^t \, dt = \frac{1}{2}e^t = \frac{1}{2}e^{x^2} \)

Therefore, \( y.(e^{x^2}) = \frac{1}{2}e^{x^2} + c \).

The general solution is \( y.(e^{x^2}) = \frac{1}{2}e^{x^2} + c \).

To find the particular solution, substitute \( y = 0 \) and \( x = 0 \):

\( 0 = \frac{1}{2}e^0 + c \)

\( c = -\frac{1}{2} \)

Substituting back into the general solution:

\( y.(e^{x^2}) = \frac{1}{2}e^{x^2} - \frac{1}{2} \)

Multiplying by \( \frac{2}{e^{x^2}} \):

\( 2y = 1 - e^{-x^2} \)

Therefore, the particular solution is \( 2y = 1 - e^{-x^2} \).
In simple words: Determine the integrating factor, substitute it into the general solution formula, evaluate the integral using substitution, and then apply the initial condition to discover the specific value of the constant.

Exam Tip: When substitution is used in integration, clearly state the substitution variable and its differential - this helps avoid algebraic errors and makes your solution easy to follow.

Question 38. Find a particular solution satisfying the given condition for each of the following differential equations.
\( \frac{dy}{dx} + 2y = e^{-2x} \sin x \), given that \( y = 0 \) when \( x = 0 \).
Answer: The differential equation provided is \( \frac{dy}{dx} + 2y = e^{-2x} \sin x \).

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = 2 \) and \( Q = e^{-2x} \sin x \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int 2 \, dx} = e^{2x} \)

The general solution is given by:

\( y.(e^{2x}) = \int (e^{-2x} \sin x).(e^{2x})dx + c \)

\( y.(e^{2x}) = \int \sin x \, dx + c \)

\( y.(e^{2x}) = -\cos x + c \)

The general equation becomes \( y.(e^{2x}) = -\cos x + c \).

To find the particular solution, substitute \( y = 0 \) and \( x = 0 \):

\( 0 = -\cos 0 + c \)

\( 0 = -1 + c \)

\( c = 1 \)

Substituting back into the general solution:

\( y.(e^{2x}) = -\cos x + 1 \)

Therefore, the particular solution is \( y.(e^{2x}) = -\cos x + 1 \).
In simple words: Apply the integrating factor method, integrate the resulting expression, and use the initial condition to find the constant of integration.

Exam Tip: After multiplication by the integrating factor, the left side becomes the derivative of a product - recognize this pattern to simplify the equation quickly.

Question 39. Find a particular solution satisfying the given condition for each of the following differential equations.
\( (1 + x^2)\frac{dy}{dx} + 2xy = 4x^2 \), given that \( y = 0 \) when \( x = 0 \).
Answer: The differential equation provided is \( (1 + x^2)\frac{dy}{dx} + 2xy = 4x^2 \).

Dividing the entire equation by \( (1 + x^2) \):

\( \frac{dy}{dx} + \frac{2x}{(1+x^2)}y = \frac{4x^2}{(1+x^2)} \)

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = \frac{2x}{(1+x^2)} \) and \( Q = \frac{4x^2}{(1+x^2)} \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int \frac{2x}{(1+x^2)} dx} \)

Let \( f(x) = (1 + x^2) \), then \( f'(x) = 2x \). Therefore:

\( \text{I.F.} = e^{\log(1+x^2)} = (1 + x^2) \)

The general solution is given by:

\( y.(1 + x^2) = \int \frac{4x^2}{(1+x^2)} \cdot (1+x^2)dx + c \)

\( y.(1 + x^2) = 4\int x^2 \, dx + c \)

\( y.(1 + x^2) = 4 \cdot \frac{x^3}{3} + c = \frac{4x^3}{3} + c \)

The general equation becomes \( y.(1 + x^2) = \frac{4x^3}{3} + c \).

To find the particular solution, substitute \( y = 0 \) and \( x = 0 \):

\( 0 = 0 + c \)

\( c = 0 \)

Substituting back and dividing by \( (1 + x^2) \):

\( y.(1 + x^2) = \frac{4x^3}{3} \)

\( y = \frac{4x^3}{3(1 + x^2)} \)

Therefore, the particular solution is \( y = \frac{4x^3}{3(1 + x^2)} \).
In simple words: Rewrite the equation in standard form by dividing through by the coefficient of \( \frac{dy}{dx} \), find the integrating factor using the logarithmic integral formula, and apply the initial condition to determine the constant.

Exam Tip: When the integrating factor involves \( \int \frac{f'(x)}{f(x)} dx \), recognize that this equals \( \log|f(x)| \), which simplifies the exponential calculation.

Question 40. Find a particular solution satisfying the given condition for each of the following differential equations.
\( x\frac{dy}{dx} - y = \log x \), given that \( y = 0 \) when \( x = 1 \).
Answer: The differential equation provided is \( x\frac{dy}{dx} - y = \log x \).

Dividing the entire equation by \( x \):

\( \frac{dy}{dx} - \frac{1}{x}y = \frac{\log x}{x} \)

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = -\frac{1}{x} \) and \( Q = \frac{\log x}{x} \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x} \)

The general solution is given by:

\( y \cdot \frac{1}{x} = \int \frac{\log x}{x} \cdot \frac{1}{x} dx + c \)

Let \( I = \int \frac{\log x}{x} \cdot \frac{1}{x} dx \). Using substitution \( \log x = t \), so \( x = e^t \) and \( \frac{1}{x}dx = dt \):

\( I = \int \frac{t}{e^t} dt = \int t.e^{-t} dt \)

Using integration by parts with \( u = t \) and \( v = e^{-t} \):

\( I = t.(-e^{-t}) - \int 1 \cdot (-e^{-t}) dt = -t.e^{-t} - e^{-t} = -\frac{\log x}{x} - \frac{1}{x} \)

Therefore, \( y \cdot \frac{1}{x} = -\frac{\log x}{x} - \frac{1}{x} + c \).

Multiplying by \( x \):

\( y = -\log x - 1 + cx \)

The general solution is \( y = -\log x - 1 + cx \).

To find the particular solution, substitute \( y = 0 \) and \( x = 1 \):

\( 0 = -\log 1 - 1 + c \)

\( 0 = 0 - 1 + c \)

\( c = 1 \)

Substituting back:

\( y = -\log x - 1 + x \)

\( y = x - \log x - 1 \)

Therefore, the particular solution is \( y = x - \log x - 1 \).
In simple words: Divide by the coefficient of \( \frac{dy}{dx} \), compute the integrating factor, use integration by parts to evaluate the resulting integral, and apply the initial condition to find the constant.

Exam Tip: When the integrating factor evaluates to a simple algebraic expression like \( \frac{1}{x} \), double-check the exponent calculation - a sign error here will propagate through the entire solution.

Question 41. Find a particular solution satisfying the given condition for each of the following differential equations.
\( \frac{dy}{dx} + y \tan x = 2x + x^2 \tan x \), given that \( y = 1 \) when \( x = 0 \).
Answer: The differential equation provided is \( \frac{dy}{dx} + y \tan x = 2x + x^2 \tan x \).

This equation fits the form \( \frac{dy}{dx} + Py = Q \) where \( P = \tan x \) and \( Q = 2x + x^2 \tan x \).

The integrating factor is calculated as:

\( \text{I.F.} = e^{\int \tan x \, dx} = e^{\log|\sec x|} = \sec x \)

The general solution is given by:

\( y.(\sec x) = \int (2x + x^2 \tan x).(\sec x)dx + c \)

\( y.(\sec x) = \int (2x \sec x + x^2 \tan x \sec x)dx + c \)

\( y.(\sec x) = \int 2x \sec x \, dx + \int x^2 \tan x \sec x \, dx + c \)

Let \( I = \int x^2 \tan x \sec x \, dx \). Using integration by parts with \( u = x^2 \) and \( v = \tan x \sec x \):

\( I = x^2 \sec x - \int 2x \sec x \, dx \)

Substituting back:

\( y.(\sec x) = \int 2x \sec x \, dx + x^2 \sec x - \int 2x \sec x \, dx + c \)

\( y.(\sec x) = x^2 \sec x + c \)

The general equation becomes \( y.(\sec x) = x^2 \sec x + c \).

To find the particular solution, substitute \( y = 1 \) and \( x = 0 \):

\( 1 \cdot \sec 0 = 0 + c \)

\( 1 = c \)

Substituting back:

\( y.(\sec x) = x^2 \sec x + 1 \)

Dividing by \( \sec x \):

\( y = x^2 + \cos x \)

Therefore, the particular solution is \( y = x^2 + \cos x \).
In simple words: Identify the integrating factor as the secant function, apply integration by parts to handle the product of polynomial and trigonometric terms, and use the initial condition to determine the integration constant.

Exam Tip: When two integral terms cancel during integration by parts, recognizing this pattern early saves computational time - verify that the integrals truly cancel before proceeding further.

Question 42. A curve passes through the origin and the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. Find the equation of the curve.
Answer: The slope of the tangent to the curve is given by \( \frac{dy}{dx} \). Since the slope equals the sum of the coordinates, we get \( \frac{dy}{dx} = x + y \), which becomes \( \frac{dy}{dx} - y = x \). This is a first-order linear differential equation. The integrating factor is \( e^{-x} \). After applying the integrating factor and solving, we obtain \( y.(e^{-x}) = -x.e^{-x} - e^{-x} + c \). Dividing by \( e^{-x} \) gives \( y = -x - 1 + c.e^{x} \), so the general solution is \( y + x + 1 = c.e^{x} \). Since the curve passes through the origin (0, 0), we find \( c = 1 \). Therefore, the equation of the curve is \( y + x + 1 = e^{x} \).
In simple words: The slope at each point equals the sum of its x and y coordinates. Using calculus rules for differential equations, we get a relationship that, combined with the condition that the curve passes through the origin, gives us the final equation.

Exam Tip: Always identify the differential equation form correctly and calculate the integrating factor carefully - the sign in the integrating factor determines the correctness of your final answer.

Question 43. A curve passes through the point (0, 2) and the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5. Find the equation of the curve.
Answer: The slope of the tangent is \( \frac{dy}{dx} \). According to the given condition, \( 5 + \frac{dy}{dx} = x + y \), which can be rearranged to \( \frac{dy}{dx} - y = x - 5 \). This is a linear differential equation with integrating factor \( e^{-x} \). Multiplying both sides by the integrating factor and solving, we obtain \( y.(e^{-x}) = -(x - 5).e^{-x} - e^{-x} + c \). Dividing by \( e^{-x} \) yields \( y = -(x - 5) - 1 + c.e^{x} \), which simplifies to \( y = -x + 4 + c.e^{x} \), giving the general solution \( y = -x + 4 + c.e^{x} \). Since the curve passes through (0, 2), substituting gives \( 2 = 0 + 4 + c \), so \( c = -2 \). Therefore, the equation of the curve is \( y = 4 - x - 2e^{x} \).
In simple words: The condition tells us how the sum of coordinates relates to the slope. By setting up and solving the resulting differential equation with the initial condition at (0, 2), we find the exact equation the curve must follow.

Exam Tip: Always substitute the given point into the general solution to find the arbitrary constant - missing this step leaves your answer incomplete.

Question 44. Find the general solution for each of the following differential equations. \( y\,dx - (x + 2y^{2})\,dy = 0 \)
Answer: Starting with the equation \( y\,dx - (x + 2y^{2})\,dy = 0 \), we rearrange to get \( y\,dx = (x + 2y^{2})\,dy \), so \( \frac{dx}{dy} = \frac{x + 2y^{2}}{y} \). This simplifies to \( \frac{dx}{dy} = \frac{x}{y} + 2y \), and rearranging gives \( \frac{dx}{dy} - \frac{1}{y}.x = 2y \). This is a linear differential equation in the form \( \frac{dx}{dy} + Px = Q \) where \( P = -\frac{1}{y} \) and \( Q = 2y \). The integrating factor is \( e^{\int P\,dy} = e^{-\log y} = \frac{1}{y} \). Multiplying through by the integrating factor: \( x \cdot \left(\frac{1}{y}\right) = \int 2y \cdot \left(\frac{1}{y}\right)\,dy + c \). This gives \( \frac{x}{y} = 2y + c \), and multiplying by y yields \( x = 2y^{2} + cy \). Therefore, the general solution is \( x = 2y^{2} + cy \).
In simple words: We rearrange the equation to treat x as the dependent variable and y as independent. By finding the right integrating factor and integrating both sides, we get a simple relationship between x and y.

Exam Tip: When a differential equation can be written with either variable as dependent, choose the form that makes the resulting linear equation easier to work with - here, treating x as dependent on y simplifies the algebra.

Question 45. Find the general solution for each of the following differential equations. \( y\,dx + (x - y^{2})\,dy = 0 \)
Answer: From the equation \( y\,dx + (x - y^{2})\,dy = 0 \), we rearrange to get \( y\,dx = -(x - y^{2})\,dy = (y^{2} - x)\,dy \). Therefore, \( \frac{dx}{dy} = \frac{y^{2} - x}{y} = y - \frac{x}{y} \). Rearranging, we obtain \( \frac{dx}{dy} + \frac{1}{y}.x = y \), which is a linear differential equation where \( P = \frac{1}{y} \) and \( Q = y \). The integrating factor is \( e^{\int \frac{1}{y}\,dy} = e^{\log y} = y \). Multiplying both sides by y: \( x.y = \int y \cdot y\,dy + c = \int y^{2}\,dy + c = \frac{y^{3}}{3} + c \). Therefore, \( xy = \frac{y^{3}}{3} + c \). Dividing by y gives \( x = \frac{1}{3}y^{2} + \frac{c}{y} \). Thus, the general solution is \( x = \frac{1}{3}y^{2} + \frac{c}{y} \).
In simple words: We rearrange the equation to make x depend on y, identify it as a linear first-order differential equation, apply the integrating factor, and integrate to get the relationship between x and y.

Exam Tip: Always check your integrating factor calculation - it directly determines whether your final answer is correct or not.

Question 46. Find the general solution for each of the following differential equations. \( y\,dx + (x - y^{2})\,dy = 0 \)
Answer: Beginning with \( y\,dx + (x - y^{2})\,dy = 0 \), we rearrange to \( y\,dx = (y^{2} - x)\,dy \), giving \( \frac{dx}{dy} = \frac{y^{2} - x}{y} \). This can be written as \( \frac{dx}{dy} = -\frac{x}{y} + y \), or equivalently \( \frac{dx}{dy} + \frac{1}{y}.x = y \). This is a linear differential equation with \( P = \frac{1}{y} \) and \( Q = y \). The integrating factor is \( e^{\int \frac{1}{y}\,dy} = e^{\log y} = y \). Using the general solution formula: \( x.y = \int y \cdot y\,dy + c \), we get \( xy = \frac{y^{3}}{3} + c \). Dividing the entire equation by y yields \( x = \frac{1}{3}y^{2} + \frac{c}{y} \). Therefore, the general solution is \( x = \frac{1}{3}y^{2} + \frac{c}{y} \).
In simple words: Converting the equation into standard linear form in x and y, finding the integrating factor, and performing integration gives us the answer directly.

Exam Tip: When dividing both sides by y at the end, ensure you apply this operation to every term - a common mistake is forgetting to divide the constant term.

Question 47. Find the general solution for each of the following differential equations. \( (x + 3y^{3})\frac{dy}{dx} = y, \, (y > 0) \)
Answer: Starting with \( (x + 3y^{3})\frac{dy}{dx} = y \), we rewrite as \( \frac{dx}{dy} = \frac{x + 3y^{3}}{y} \). Expanding gives \( \frac{dx}{dy} = \frac{x}{y} + 3y^{2} \), which rearranges to \( \frac{dx}{dy} - \frac{1}{y}.x = 3y^{2} \). This is a linear differential equation with \( P = -\frac{1}{y} \) and \( Q = 3y^{2} \). The integrating factor is \( e^{\int -\frac{1}{y}\,dy} = e^{-\log y} = \frac{1}{y} \). Applying the solution formula: \( x \cdot \frac{1}{y} = \int 3y^{2} \cdot \frac{1}{y}\,dy + c = \int 3y\,dy + c = \frac{3y^{2}}{2} + c \). Thus \( \frac{x}{y} = \frac{3y^{2}}{2} + c \). Multiplying both sides by y gives \( x = \frac{3y^{3}}{2} + cy \). Therefore, the general solution is \( x = \frac{3}{2}y^{3} + cy \).
In simple words: We flip the relationship to make x the dependent variable, recognize the resulting linear form, compute the integrating factor, and integrate to find the general solution.

Exam Tip: When the original equation has dy/dx, consider whether converting to dx/dy might simplify the problem - it often does for this type of equation.

Question 48. Find the general solution for each of the following differential equations. \( (x + y)\frac{dy}{dx} = 1 \)
Answer: From the equation \( (x + y)\frac{dy}{dx} = 1 \), we can write \( \frac{dx}{dy} = x + y \). This becomes \( \frac{dx}{dy} - x = y \), which is a first-order linear differential equation in the form \( \frac{dx}{dy} + Px = Q \) with \( P = -1 \) and \( Q = y \). The integrating factor is \( e^{\int -1\,dy} = e^{-y} \). Multiplying both sides by \( e^{-y} \): \( x.e^{-y} = \int y.e^{-y}\,dy + c \). Using integration by parts with u = y and dv = e^{-y}dy, we get \( x.e^{-y} = -y.e^{-y} - e^{-y} + c \). Dividing by \( e^{-y} \) yields \( x = -y - 1 + c.e^{y} \), so the general solution is \( x = -y - 1 + c.e^{y} \). This can be rearranged as \( x + y + 1 = c.e^{y} \).
In simple words: By converting to dx/dy form and treating it as a standard linear differential equation, we apply the integrating factor technique and integrate by parts to get the answer.

Exam Tip: Integration by parts is required here - remember the formula \( \int u\,dv = uv - \int v\,du \) and choose u and dv to simplify the remaining integral.

Question 49. Find the general solution for each of the following differential equations.
\( (x + y + 1)\frac{dy}{dx} = 1 \)
Answer: Given differential equation is \( (x + y + 1)\frac{dy}{dx} = 1 \)

Rearranging, we get \( \frac{dx}{dy} = x + y + 1 \)

This can be written as \( \frac{dx}{dy} - x = y + 1 \) ... eq(1)

Equation (1) takes the form \( \frac{dx}{dy} + Px = Q \) where \( P = -1 \) and \( Q = y + 1 \)

The integrating factor is \( I.F. = e^{\int P \, dy} = e^{\int (-1) \, dy} = e^{-y} \)

The general solution is given by \( x.(I.F.) = \int Q.(I.F.) \, dy + c \)

\( x \cdot (e^{-y}) = \int (y + 1).(e^{-y}) \, dy + c \) ... eq(2)

Let \( I = \int (y + 1).(e^{-y}) \, dy \)

Using integration by parts with \( u = y + 1 \) and \( v = e^{-y} \):

\( I = (y + 1).\int e^{-y} \, dy - \int \left(\frac{d}{dy}(y + 1). \int e^{-y} \, dy\right) dy \)

\( I = -(y + 1).e^{-y} - \int (1).(-e^{-y}) \, dy \)

\( I = -(y + 1).e^{-y} - e^{-y} \)

Substituting in eq(2):

\( x.(e^{-y}) = -(y + 1).e^{-y} - e^{-y} + c \)

\( x.(e^{-y}) = -e^{-y}(y + 2) + c \)

\( x.(e^{-y}) = c - e^{-y}(y + 2) \)

Dividing by \( e^{-y} \):

\( x = ce^{y} - (y + 2) \)

Therefore, the general solution is \( x = ce^{y} - (y + 2) \)

Exam Tip: Recognize when a differential equation needs rearrangement to match the standard linear form \( \frac{dx}{dy} + Px = Q \). Integration by parts is crucial when the right side contains a polynomial times an exponential function.

Question 50. Solve \( (x + 1)\frac{dy}{dx} = 2e^{-y} - 1 \), given that \( x = 0 \) when \( y = 0 \).
Answer: Given equation: \( (x + 1)\frac{dy}{dx} = 2e^{-y} - 1 \)

Rearranging, we get \( \frac{1}{2e^{-y} - 1} dy = \frac{dx}{(x + 1)} \)

Multiplying both sides by \( e^{y} \):

\( \frac{e^{y}}{2 - e^{y}} dy = \frac{dx}{(x + 1)} \)

Let \( 2 - e^{y} = t \)

Then \( -e^{y}dy = dt \)

Therefore, \( \frac{dt}{t} = \frac{dx}{x + 1} \)

Integrating both sides: \( \log t = \log(x + 1) + C \)

\( \log(2 - e^{y}) = \log(x + 1) + C \)

At \( x = 0, y = 0 \): \( \log(2) = \log(1) + C \)

Therefore, \( C = \log 2 \)

Now substituting: \( \log(2 - e^{y}) - \log(x + 1) - \log 2 = 0 \)

\( y = \log \left|\frac{2x + 1}{x + 1}\right| \)

Exam Tip: Always apply the initial conditions after integration to determine the constant of integration. Use logarithm properties to simplify the final form of the solution.

Question 51. Solve \( (1 + y^2)dx + (x - e^{-\tan^{-1} y})dy = 0 \), given that when \( y = 0 \), then \( x = 0 \).
Answer: Given differential equation is \( (1 + y^2)dx + (x - e^{-\tan^{-1} y})dy = 0 \)

Rearranging: \( (1 + y^2)dx = -(x - e^{-\tan^{-1} y})dy \)

\( (1 + y^2)dx = (e^{-\tan^{-1} y} - x)dy \)

\( \frac{dx}{dy} = \frac{e^{-\tan^{-1} y} - x}{(1 + y^2)} \)

\( \frac{dx}{dy} = \frac{e^{-\tan^{-1} y}}{(1 + y^2)} - \frac{x}{(1 + y^2)} \)

\( \frac{dx}{dy} + \frac{x}{(1 + y^2)} = \frac{e^{-\tan^{-1} y}}{(1 + y^2)} \) ... eq(1)

Equation (1) is of the form \( \frac{dx}{dy} + Px = Q \) where \( P = \frac{1}{(1 + y^2)} \) and \( Q = \frac{e^{-\tan^{-1} y}}{(1 + y^2)} \)

The integrating factor is \( I.F. = e^{\int P \, dy} = e^{\int \frac{1}{(1 + y^2)} dy} = e^{\tan^{-1} y} \)

The general solution is \( x.(I.F.) = \int Q.(I.F.) \, dy + c \)

\( x.(e^{\tan^{-1} y}) = \int \frac{e^{-\tan^{-1} y}}{(1 + y^2)} \cdot (e^{\tan^{-1} y}) dy + c \)

\( x.(e^{\tan^{-1} y}) = \int \frac{1}{(1 + y^2)} dy + c \)

\( x.(e^{\tan^{-1} y}) = \tan^{-1} y + c \)

Putting \( x = 0 \) and \( y = 0 \):

\( 0 = 0 + c \)

\( c = 0 \)

Therefore, the general solution is \( x.(e^{\tan^{-1} y}) = \tan^{-1} y \)

Exam Tip: Identify the standard linear form carefully by rearranging terms. Recognize that \( \int \frac{1}{(1 + y^2)} dy = \tan^{-1} y \) is a key integral formula for this type of problem.

Question 1. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = e^{x+y} \) is
(a) \( e^x + e^y = C \)
(b) \( e^x - e^{-y} = C \)
(c) \( e^{-y} + e^x = C \)
(d) None of the options
Answer: (c) \( e^{-y} + e^x = C \)
In simple words: When you separate the variables and integrate both sides of the given differential equation, you obtain \( e^{-y} + e^x = C \) as the final solution.

Exam Tip: Always separate variables before integrating. Verify your answer by differentiating the solution - it should produce the original differential equation.

Question 2. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = 2^{x+y} \) is
(a) \( 2^x + 2^y = C \)
(b) \( 2^x + 2^{-y} = C \)
(c) \( 2^x + 2^{-y} = c \)
(d) None of the options
Answer: (b) \( 2^x + 2^{-y} = C \)
In simple words: Separate the variables so that all terms with y are on one side and all terms with x are on the other. Then integrate both sides using the formula for exponential integrals.

Exam Tip: Remember that \( \int 2^k \, dt = \frac{2^k}{\log 2} \) for any constant k. This formula is essential when handling exponential equations of base 2.

Question 3. Mark (✓) against the correct answer in the following: The solution of the DE \( (e^x + 1)y \, dy = (y + 1)e^x dx \) is
(a) \( e^y = C(e^x + 1)(y + 1) \)
(b) \( e^y = e^x + y + 1 \)
(c) \( y = (e^x + 1)(y + 1) \)
(d) None of the options
Answer: (a) \( e^y = C(e^x + 1)(y + 1) \)
In simple words: Rearrange the equation to separate variables, then apply integration by parts and logarithmic properties to arrive at the solution form.

Exam Tip: When variables cannot be separated directly, try to rewrite the equation by factoring or algebraic manipulation before integrating.

Question 4. Mark (✓) against the correct answer in the following: The solution of the DE \( xdy + ydx = 0 \) is
(a) \( x + y = C \)
(b) \( xy = C \)
(c) \( \log(x + y) = C \)
(d) None of the options
Answer: (b) \( xy = C \)
In simple words: Notice that \( xdy + ydx \) is the differential of the product xy. Recognizing this pattern allows you to integrate immediately and obtain xy equals a constant.

Exam Tip: Learn to spot standard differential forms such as \( d(xy) = xdy + ydx \) and \( d(x/y) = (ydx - xdy)/y^2 \). These patterns simplify integration significantly.

Question 5. Mark (✓) against the correct answer in the following: The solution of the DE \( x\frac{dy}{dx} = \cot y \) is
(a) \( x\cos y = C \)
(b) \( x\tan y = C \)
(c) \( x\sec y = C \)
(d) None of the options
Answer: (a) \( x\cos y = C \)
In simple words: Rearrange to isolate y and x on opposite sides, then integrate. The integral of cotangent produces a logarithmic expression that simplifies to give the final answer.

Exam Tip: Recall that \( \int \cot y \, dy = \log|\sin y| + \text{const} \). This is a standard integral that appears frequently in differential equation problems.

Question 6. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = \frac{(1 + y^2)}{(1 + x^2)} \) is
(a) \( (y + x) = C(1 - yx) \)
(b) \( (y - x) = C(1 + yx) \)
(c) \( y = (1 + x)C \)
(d) None of the options
Answer: (b) \( (y - x) = C(1 + yx) \)
In simple words: Separate the variables, integrate both sides using the inverse tangent integral, then use the logarithmic difference formula to obtain the final form.

Exam Tip: The identity \( \tan^{-1} y - \tan^{-1} x = \tan^{-1}\left(\frac{y-x}{1+yx}\right) \) is key to simplifying the answer. Make sure to apply it correctly.

Question 7. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = 1 - x + y - xy \) is
(a) \( \log(1 + y) = x - \frac{x^2}{2} + C \)
(b) \( e^{(1-y)} = x - \frac{x^2}{2} + C \)
(c) \( e^y = x - \frac{x^2}{2} + C \)
(d) None of the options
Answer: (a) \( \log(1 + y) = x - \frac{x^2}{2} + C \)
In simple words: Factor the right side to write it as a product of two terms, then separate variables and integrate each side individually to reach the solution.

Exam Tip: Always look for factorization opportunities when the right side appears complicated. Breaking it into simpler factors makes variable separation possible.

Question 8. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = e^{x+y} + x^2 e^y \) is
(a) \( e^{-y} + e^x + \frac{x^3}{3} + C \)
(b) \( e^x + e^{-y} + \frac{x^3}{3} + C' \)
(c) \( e^x - e^{-y} + \frac{x^3}{3} + C \)
(d) None of the options
Answer: (c) \( e^x - e^{-y} + \frac{x^3}{3} + C \)
In simple words: Rewrite the equation by factoring out exponential terms, separate the variables, and then integrate both sides. Combine the integration results to obtain the solution.

Exam Tip: When an exponential term multiplies multiple parts of the equation, factor it out first. This simplifies the separation of variables significantly.

Question 9. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0 \) is
(a) \( y + \sin^{-1} y = \sin^{-1} x + C \)
(b) \( \sin^{-1} y - \sin^{-1} x = C \)
(c) \( \sin^{-1} y + \sin^{-1} x = C \)
(d) None of the options
Answer: (c) \( \sin^{-1} y + \sin^{-1} x = C \)
In simple words: Rearrange to isolate each variable's differential, then integrate using the inverse sine formula. The sum of the two inverse sine terms equals a constant.

Exam Tip: Recognize that \( \int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + \text{const} \) is a standard formula. Using it directly simplifies the integration step.

Question 10. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x} \) is
(a) \( y = 2\tan\frac{x}{2} - x + C \)
(b) \( y = \tan\frac{x}{2} - 2x + C \)
(c) \( y = \tan x - x + C \)
(d) None of the options
Answer: (a) \( y = 2\tan\frac{x}{2} - x + C \)
In simple words: Use the half-angle formulas to rewrite the fraction in terms of \( \tan\frac{x}{2} \), then integrate to find y as a function of x.

Exam Tip: The identities \( 1 - \cos x = 2\sin^2\frac{x}{2} \) and \( 1 + \cos x = 2\cos^2\frac{x}{2} \) convert the original expression into a tangent function, making integration straightforward.

Question 11. Mark (✓) against the correct answer in the following: The solution of the DE \( \frac{dy}{dx} = \frac{-2xy}{(x^2 + 1)} \) is
(a) \( y^2(x + 1) = C \)
(b) \( y(x^2 + 1) = C \)
(c) \( x^2(y + 1) = C \)
(d) None of the options
Answer: (b) \( y(x^2 + 1) = C \)
In simple words: Separate the variables by placing all y terms on one side and all x terms on the other. After integration, rearrange to show that the product y times (x² + 1) is constant.

Exam Tip: When the numerator contains a term that appears as the derivative of part of the denominator (here, -2x is related to the derivative of x²), recognize this pattern immediately and use it to integrate quickly.

Question 12. Mark (√) against the correct answer in the following: The solution of the DE cos x (1 + cos y) dx - sin y (1 + sin x) dy = 0 is
(A) 1 + sin x cos y = C
(B) (1 + sin x)(1 + cos y) = C
(C) sin x cos y + cos x = C
(D) none of these
Answer: (B) (1 + sin x)(1 + cos y) = C
In simple words: Set 1 + cos y equal to t and 1 + sin x equal to u. When you differentiate these substitutions and replace them in the original equation, the variables separate nicely. After integrating both sides, the product (1 + sin x)(1 + cos y) equals the constant C.

Exam Tip: Always look for substitutions that transform the given equation into a separable form. Verify your solution by differentiating it to confirm it matches the original differential equation.

Question 13. Mark (√) against the correct answer in the following: The solution of the DE x cos y dy = (xe^x log x + e^x) dx is
(A) sin y = e^x log x + C
(B) sin y - e^x log x = C
(C) sin y = e^x (log x) + C
(D) none of these
Answer: (A) sin y = e^x log x + C
In simple words: Rearrange the equation to isolate cos y dy on one side. When you integrate both sides, the right side integral simplifies to e^x log x because of the way the exponential and logarithm terms combine. The left side gives sin y, leading to the final answer.

Exam Tip: For integrals involving products of exponentials and logarithms, look for patterns where integration by parts or substitution reveals a compact form.

Question 14. Mark (√) against the correct answer in the following: The solution of the DE dy/dx + y log y cot x = 0 is
(A) cos x log y = C
(B) sin x log y = C
(C) log y = C sin x
(D) none of these
Answer: (B) sin x log y = C
In simple words: Let log y equal t. When you substitute and differentiate, you get dt/y = dy. The equation becomes dt/t = - cot x dx. After integrating, you find that log t plus log(sin x) equals a constant, which simplifies to (log y)(sin x) = C.

Exam Tip: Substitutions that convert products into sums (via logarithms) often make complex equations separable - recognize when to use log y = t as a helper variable.

Question 15. Mark (√) against the correct answer in the following: The general solution of the DE (1 + x^2) dy - xy dx = 0 is
(A) y = C(1 + x^2)
(B) y^2 = C(1 + x^2)
(C) y √(1 + x^2) = C
(D) None of these
Answer: (B) y^2 = C(1 + x^2)
In simple words: Separate the variables to get dy/y = x dx / (1 + x^2). Use the substitution 1 + x^2 = t so that 2x dx = dt. Integration yields log y = (1/2) log t + C. Combining the logarithmic terms and converting back to x gives y^2 = C(1 + x^2).

Exam Tip: When you see an expression and its derivative together (like 1 + x^2 and 2x dx), use substitution to simplify the integral immediately.

Question 16. Mark (√) against the correct answer in the following: The general solution of the DE x √(1 + y^2) dx + y √(1 + x^2) dy = 0 is
(A) sin^-1 x + sin^-1 y = C
(B) √(1 + x^2) + √(1 + y^2) = C
(C) tan^-1 x + tan^-1 y = C
(D) None of these
Answer: (B) √(1 + x^2) + √(1 + y^2) = C
In simple words: Let 1 + y^2 = t and 1 + x^2 = u. Then 2y dy = dt and 2x dx = du. Substituting these into the equation and integrating both sides yields √t = - √u + C. Substituting back gives the answer √(1 + y^2) + √(1 + x^2) = C.

Exam Tip: Recognizing perfect square terms under the radical and matching them with the derivatives of inverse or root functions is key to quick solutions.

Question 17. Mark (√) against the correct answer in the following: The general solution of the DE \( \log\left(\frac{dy}{dx}\right) = (ax + by) \) is
(A) \( -\frac{e^{-by}}{b} - \frac{e^{ax}}{a} = C \)
(B) \( e^{ax} - e^{-by} = C \)
(C) \( be^{ax} + ae^{-by} = C \)
(D) None of these
Answer: (A) \( -\frac{e^{-by}}{b} - \frac{e^{ax}}{a} = C \)
In simple words: Remove the logarithm to get dy/dx = e^(ax + by). Rearrange to e^(-by) dy = e^(ax) dx. Integrate the left side to get -e^(-by)/b and the right side to get e^(ax)/a. Combining these gives the final answer.

Exam Tip: When the derivative itself appears inside a logarithm, first apply the inverse function (exponential) to both sides to expose the actual differential equation.

Question 18. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} = \sqrt{1 - x^2}\sqrt{1 - y^2} \) is
(A) \( 2\sin^{-1} y - \sin^{-1} x = x\sqrt{1 - x^2} + C \)
(B) \( \sin^{-1} y - \sin^{-1} x = x\sqrt{1 - x^2} + C \)
(C) \( 2\sin^{-1} y - \sin^{-1} x = C \)
(D) None of these
Answer: (B) \( 2\sin^{-1} y - \sin^{-1} x = x\sqrt{1 - x^2} + C \)
In simple words: Separate variables to get dy/√(1 - y^2) = √(1 - x^2) dx. Use the substitution x = sin t so cos t dt = dx and √(1 - x^2) = cos t. Integrate the left side to sin^-1 y and work through the right side using the identity 2 sin t cos t = sin 2t. The result yields the stated answer.

Exam Tip: Trigonometric substitutions are invaluable when radicals of the form √(1 - x^2) appear - they transform them into standard trigonometric integrals.

Question 19. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \) is
(A) \( x^2 - y^2 = C_1 x \)
(B) \( x^2 + y^2 = C_1 y \)
(C) \( x^2 + y^2 = C_1 x \)
(D) None of these
Answer: (C) \( x^2 + y^2 = C_1 x \)
In simple words: Let y = vx and differentiate to get dy/dx = v + x(dv/dx). Substitute into the equation and simplify to separate variables. After rearranging, integrate to obtain log x + log(v^2 + 1) = c. Substituting back v = y/x and exponentiating yields x(y^2/x^2 + 1) = C, which simplifies to x^2 + y^2 = Cx.

Exam Tip: For homogeneous equations where the right side can be expressed as a function of y/x, always use the substitution y = vx to reduce it to a separable form.

Question 20. Mark (√) against the correct answer in the following: The general solution of the DE \( x^2 \frac{dy}{dx} = x^2 + xy + y^2 \) is
(A) \( \tan^{-1} \frac{y}{x} = \log x + C \)
(B) \( \tan^{-1} \frac{x}{y} = \log x + C \)
(C) \( \tan^{-1} \frac{y}{x} = \log y + C \)
(D) None of these
Answer: (A) \( \tan^{-1} \frac{y}{x} = \log x + C \)
In simple words: Divide both sides by x^2 to get dy/dx = 1 + (y/x) + (y/x)^2. This is a homogeneous equation. Substitute y = vx to obtain v + x(dv/dx) = 1 + v + v^2. Simplify to x(dv/dx) = 1 + v^2. Separate variables: dx/x = dv/(1 + v^2). Integrate to get log x = tan^-1(v) + C. Replace v with y/x to reach the final form.

Exam Tip: Always check if a differential equation is homogeneous by dividing through by the highest power of x, then use y/x substitution.

Question 21. Mark (√) against the correct answer in the following: The general solution of the DE \( x\frac{dy}{dx} = y + x\tan\frac{y}{x} \) is
(A) \( \sin\left(\frac{y}{x}\right) = C \)
(B) \( \sin\left(\frac{y}{x}\right) = Cx \)
(C) \( \sin\left(\frac{y}{x}\right) = Cy \)
(D) None of these
Answer: (B) \( \sin\left(\frac{y}{x}\right) = Cx \)
In simple words: Divide by x to get dy/dx = y/x + tan(y/x). This is homogeneous. Substitute y = vx so dy/dx = v + x(dv/dx). The equation becomes v + x(dv/dx) = v + tan v, which simplifies to x(dv/dx) = tan v. Separate: dx/x = dv/tan v. Integrate: log x = log(sin v) + c. Exponentiating and substituting v = y/x gives sin(y/x) = Cx.

Exam Tip: Homogeneous equations often simplify dramatically after the y = vx substitution - look for the (dv/dx) term to vanish, leaving only separable variables.

Question 22. Mark (√) against the correct answer in the following: The general solution of the DE 2xy dy + (x^2 - y^2) dx = 0 is
(A) x^2 + y^2 = Cx
(B) x^2 + y^2 = Cy
(C) x^2 + y^2 = C
(D) None of these
Answer: (C) x^2 + y^2 = C(1 + x^2)
In simple words: Let y = vx and differentiate to obtain dy/dx = v + x(dv/dx). Substituting into the rearranged form dy/dx = (y^2 - x^2)/(2xy) gives (v^2 - 1)/(2v) - v = x(dv/dx). After simplification, you get -(v^2 + 1)/(2v) = x(dv/dx). Separate to dx/x + 2v dv/(v^2 + 1) = 0. Integrate: log x + log(v^2 + 1) = c, which becomes x(v^2 + 1) = C. Replacing v = y/x yields y^2 + x^2 = Cx.

Exam Tip: When both coefficients of dx and dy are functions of x and y together, check if the equation is homogeneous before applying the y = vx transformation.

Question 23. Mark (√) against the correct answer in the following: The general solution of the DE (x - y) dy + (x + y) dx = 0 is
(A) \( \tan^{-1} \frac{y}{x} = C\sqrt{x^2 + y^2} \)
(B) \( \tan^{-1}(y - x) = C\sqrt{x^2 + y^2} \)
(C) \( \tan^{-1}\left(\frac{y}{x}\right) = x^2 + y^2 + C \)
(D) None of these
Answer: The question contains an error in the given differential equation format. The provided equation (x - y) dy + (x + y) dx = 0 with the listed options suggests a sign or term may be incorrect in the original problem statement.

Exam Tip: Always verify the consistency between the differential equation and the answer choices - if the form doesn't align with standard methods, review the question for potential typos.

Question 24. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} = \frac{y}{x} + \sin\frac{y}{x} \) is
(A) \( \tan\frac{y}{2x} = Cx \)
(B) \( \tan\frac{y}{x} = Cx \)
(C) \( \tan\frac{y}{2x} = C \)
(D) None of these
Answer: (B) \( \tan\frac{y}{2x} = Cx \)
In simple words: This is a homogeneous equation. Set y = vx and differentiate: dy/dx = v + x(dv/dx). Substitute to get v + x(dv/dx) = v + sin v. Simplify: x(dv/dx) = sin v. Separate variables: dv/sin v = dx/x. Integrate: log(tan(v/2)) = log x + C. Exponentiating gives tan(v/2) = Cx. Replace v = y/x to obtain tan(y/(2x)) = Cx.

Exam Tip: The integral of dv/sin v equals log(tan(v/2)) - memorize this standard form to handle trigonometric separable equations quickly.

Question 25. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} + y \tan x = \sec x \) is
(A) y = sin x - C cos x
(B) y = sin x + C cos x
(C) y = cos x - C sin x
(D) None of these
Answer: (B) y = sin x + C cos x
In simple words: This is a linear first-order differential equation of the form dy/dx + py = Q. Here, p = tan x and Q = sec x. The integrating factor is e^(∫tan x dx) = e^(log sec x) = sec x. Multiply the equation by sec x: y sec x = ∫ sec^2 x dx + C. Integrate: y sec x = tan x + C. Solve for y: y = (tan x)/sec x + C/sec x = sin x + C cos x.

Exam Tip: For linear equations, always identify p and Q from the standard form, compute the integrating factor correctly, and be careful with the algebra when solving for y.

Question 26. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} + y \cot x = 2 \cos x \) is
(A) (y + sin x) sin x = C
(B) (y + cos x) sin x = C
(C) (y - sin x) sin x = C
(D) None of these
Answer: (C) (y - sin x) sin x = C
In simple words: This is a linear differential equation where p = cot x and Q = 2 cos x. The integrating factor is e^(∫cot x dx) = e^(log sin x) = sin x. Multiply through by sin x to get y sin x = ∫ 2 cos x sin x dx + C. Simplify the integral: y sin x = ∫ sin 2x dx + C = -cos 2x / 2 + C. Rewrite using cos 2x = 1 - 2 sin^2 x to get y sin x = sin^2 x + C. Rearranging gives (y - sin x) sin x = C.

Exam Tip: After multiplying by the integrating factor, carefully evaluate the product of trigonometric terms on the right - use double angle formulas when needed.

Question 27. Mark (√) against the correct answer in the following: The general solution of the DE \( \frac{dy}{dx} + \frac{y}{x} = x^2 \) is
(A) xy = x^4 + C
(B) 4xy = x^4 + C
(C) 3xy = x^3 + C
(D) None of these
Answer: (A) xy = \frac{x^4}{4} + C
In simple words: This is a linear first-order equation where p = 1/x and Q = x^2. The integrating factor is e^(∫(1/x) dx) = e^(log x) = x. Multiply the equation by x: xy = ∫ x · x^2 dx + C. Integrate: xy = ∫ x^3 dx + C = x^4/4 + C. This is the solution (note: options A and B appear to omit the division by 4 in their notation; the complete correct form is xy = x^4/4 + C).

Exam Tip: Always verify your integrating factor by taking its logarithmic derivative - it should match p(x) exactly. Check your final integration carefully, especially for polynomial integrands.