RS Aggarwal Solutions for Class 12 Chapter 19 Differential Equations with Variable Separable

Exercise 19A

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

Rearranging the terms, we get:
\( \frac{dy}{1 + y^2} = (1 + x^2)dx \)

Integrating both sides, we get:
\( \int \frac{dy}{1 + y^2} = \int (1 + x^2)dx + c \)

\( \tan^{-1} y = x + \frac{x^3}{3} + c \)

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 each side separately. The integral of 1/(1 + y²) gives you the inverse tangent of y, and the integral of (1 + x²) gives you x plus x³/3.

Exam Tip: Always separate variables completely before integrating. Remember that \( \int \frac{dy}{1 + y^2} = \tan^{-1} y \) and \( \int x^n dx = \frac{x^{n+1}}{n+1} \).

Question 2. Find the general solution of each of the following differential equations: \( x^4 \frac{dy}{dx} = -y^4 \)
Answer: We have \( x^4 \frac{dy}{dx} = -y^4 \).

Rearranging the terms, we get:
\( \frac{dy}{-y^4} = \frac{dx}{x^4} \)

Integrating both sides, we get:
\( \int \frac{dy}{-y^4} = \int \frac{dx}{x^4} + c' \)

\( \frac{-y^{-4 + 1}}{-4 + 1} = \frac{x^{-4 + 1}}{-4 + 1} + c' \)

\( \frac{1}{3y^3} = -\frac{1}{3x^3} + c' \)

\( \frac{1}{y^3} + \frac{1}{x^3} = 3c' = c \)

In simple words: Move the variables to opposite sides of the equation so y terms are with dy and x terms are with dx. When you integrate negative powers, remember to apply the power rule: \( \int x^n dx = \frac{x^{n+1}}{n+1} \). Simplify the fractions after integration.

Exam Tip: Be careful with negative exponents - use the power rule correctly. Combine the two reciprocal cubic terms into a single equation and simplify the constant at the end.

Question 3. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = 1 + x + y + xy \)
Answer: We have \( \frac{dy}{dx} = 1 + x + y + xy = 1 + y + x(1 + y) = (1 + y)(1 + x) \).

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

Integrating both sides, we get:
\( \int \frac{dy}{1 + y} = \int (1 + x)dx + c \)

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

In simple words: Start by factoring the right side to separate the variables. Group terms containing y together and terms with x together. Once separated, integrate the natural logarithm on the left and the polynomial on the right.

Exam Tip: Always look for opportunities to factor the right side of the differential equation. This makes it easier to separate variables. Recognize that \( \int \frac{dy}{1 + y} = \log|1 + y| \).

Question 4. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = 1 - x + y - xy \)
Answer: We have \( \frac{dy}{dx} = 1 - x + y - xy = 1 + y - x(1 + y) = (1 + y)(1 - x) \).

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

Integrating both sides, we get:
\( \int \frac{dy}{1 + y} = \int (1 - x)dx + c \)

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

In simple words: Factor the expression on the right by grouping the y-dependent terms and x-dependent terms separately. This allows you to separate the variables. Then integrate each side using the standard formulas for logarithm and polynomial integrals.

Exam Tip: Watch the signs carefully when factoring - notice that (1 - x) has a minus sign. After integrating, double-check that the coefficient of x² is negative on the right side.

Question 5. Find the general solution of each of the following differential equations: \( (x - 1)\frac{dy}{dx} = 2x^3 y \)
Answer: We have \( (x - 1)\frac{dy}{dx} = 2x^3 y \).

Separating the variables, we get:
\( \frac{dy}{y} = \frac{2x^3}{x - 1}dx \)

We can rewrite the right side as:
\( \frac{dy}{y} = \frac{2[(x - 1)(x^2 + x + 1) + 1]}{x - 1}dx = 2\left(x^2 + x + 1 + \frac{1}{x - 1}\right)dx \)

Integrating both sides, we get:
\( \int \frac{dy}{y} = \int 2\left(x^2 + x + 1 + \frac{1}{x - 1}\right)dx + c \)

\( \log|y| = \frac{2x^3}{3} + x^2 + 2x + 2\log|x - 1| + c \)

In simple words: Separate variables by moving all y terms to the left and all x terms to the right. The right side needs polynomial division - divide 2x³ by (x - 1) to get a quotient plus a remainder. Then integrate both sides using standard formulas.

Exam Tip: When the integrand is a rational function with a degree in the numerator greater than or equal to the denominator, use polynomial long division first. This breaks the integral into simpler terms.

Question 6. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = e^{x + y} \)
Answer: We have \( \frac{dy}{dx} = e^{x + y} = e^x \cdot e^y \).

Rearranging the terms, we get:
\( \frac{dy}{e^y} = e^x dx \)

Integrating both sides, we get:
\( \int \frac{dy}{e^y} = \int e^x dx + c \)

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

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

In simple words: Use the property that e^(x + y) equals e^x times e^y to separate the exponential into two parts. Move the e^y term to the other side, then integrate. Remember that the integral of e^x is e^x and the integral of e^(-y) with respect to y is -e^(-y).

Exam Tip: Always split exponentials with sums in the exponent using the product rule. Pay close attention to negative signs when integrating e^(-y).

Question 7. Find the general solution of each of the following differential equations: \( (e^x + e^{-x})dy - (e^x - e^{-x})dx = 0 \)
Answer: We have \( (e^x + e^{-x})dy - (e^x - e^{-x})dx = 0 \).

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

Integrating both sides, we get:
\( \int dy = \int \frac{e^x - e^{-x}}{e^x + e^{-x}}dx + c \)

Notice that the numerator is the derivative of the denominator. Using the substitution property:
\( y = \log|e^x + e^{-x}| + c \)

In simple words: Rearrange to separate variables. Notice that the numerator on the right is exactly the derivative of the denominator. When this pattern appears (a fraction where the top is the derivative of the bottom), the integral of that fraction is the natural logarithm of the denominator.

Exam Tip: Watch for the pattern where \( \int \frac{f'(x)}{f(x)}dx = \log|f(x)| \). This is a powerful shortcut and saves time on integration.

Question 8. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = e^x e^{-y} + x^2 e^{-y} \)
Answer: Given: \( \frac{dy}{dx} = e^x e^{-y} + x^2 e^{-y} = e^{-y}(e^x + x^2) \).

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

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

Rearranging, we get:
\( e^y dy = (e^x + x^2)dx \)

Integrating both sides, we get:
\( \int e^y dy = \int (e^x + x^2)dx + c \)

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

In simple words: Factor out the e^(-y) term from the right side. Then move e^(-y) to the left side by multiplying both sides by e^y. This flips e^(-y) to e^y on the left. Now integrate both sides using the fact that the integral of e^y is e^y.

Exam Tip: When you see e^(-y) multiplying something on the right, multiply both sides by e^y to get e^y on the left. This makes the integration straightforward.

Question 9. Find the general solution of each of the following differential equations: \( e^{2x - 3y}dx + e^{2y - 3x}dy = 0 \)
Answer: We have \( e^{2x - 3y}dx + e^{2y - 3x}dy = 0 \).

This can be rewritten as:
\( e^{2x} e^{-3y}dx + e^{2y} e^{-3x}dy = 0 \)

Rearranging the terms, we get:
\( e^{2x + 3x}dx = -e^{2y + 3y}dy \)
\( e^{5x}dx = -e^{5y}dy \)

Integrating both sides, we get:
\( \int e^{5x}dx = -\int e^{5y}dy + c' \)

\( \frac{e^{5x}}{5} = -\frac{e^{5y}}{5} + c' \)
\( e^{5x} + e^{5y} = 5c' = c \)

In simple words: Split each exponential using exponent rules so that e^(2x) is separate from e^(-3y). Rearrange to group all x terms on one side and all y terms on the other. When you integrate e^(5x), remember to divide by the coefficient 5 in the exponent.

Exam Tip: When integrating \( e^{ax} \), the result is \( \frac{e^{ax}}{a} \). Don't forget this coefficient - it's a common source of errors.

Question 10. Find the general solution of each of the following differential equations: \( e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0 \)
Answer: We have \( e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0 \).

Rearranging all the terms, we get:
\( \frac{e^x dx}{1 - e^x} = -\frac{\sec^2 y \, dy}{\tan y} \)

Integrating both sides, we get:
\( \int \frac{e^x dx}{1 - e^x} = -\int \frac{\sec^2 y \, dy}{\tan y} + c \)

\( \frac{\log|1 - e^x|}{-1} = -\log|\tan y| + \log c \)

\( -\log|1 - e^x| = -\log|\tan y| + \log c \)
\( \log|1 - e^x| + \log c = \log|\tan y| \)
\( \tan y = c(1 - e^x) \)

In simple words: Separate the variables by moving all x terms and their dx to one side and all y terms with dy to the other. Notice that when you integrate \( \sec^2 y / \tan y \), you get the logarithm of the tangent. Use logarithm properties to combine and simplify.

Exam Tip: Remember that the integral of sec² y is tan y, and the integral of 1/tan y is log|tan y|. Logarithm rules allow you to move constants in and out of the log, which helps simplify the final answer.

Question 11. Find the general solution of each of the following differential equations: \( \sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 \)
Answer: We have \( \sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0 \).

Rearranging the terms, we get:
\( \frac{\sec^2 x \, dx}{\tan x} = -\frac{\sec^2 y \, dy}{\tan y} \)

Integrating both sides, we get:
\( \int \frac{\sec^2 x \, dx}{\tan x} = -\int \frac{\sec^2 y \, dy}{\tan y} + c \)

\( \log|\tan x| = -\log|\tan y| + \log c \)
\( \log|\tan x| + \log|\tan y| = \log c \)
\( \tan x \cdot \tan y = c \)

In simple words: Move sec² x and tan y to opposite sides of the differential equation. The key insight is that sec²(x)/tan(x) integrates to log|tan(x)|. After integrating both sides, use logarithm rules to combine the logs on one side.

Exam Tip: This problem uses the formula \( \int \frac{\sec^2 \theta}{\tan \theta} d\theta = \log|\tan \theta| \). Recognize this pattern and apply it directly. The final answer is a product of two trigonometric functions equal to a constant.

Question 12. Find the general solution of each of the following differential equations: \( \cos x(1 + \cos y)dx - \sin y(1 + \sin x)dy = 0 \)
Answer: We have \( \cos x(1 + \cos y)dx - \sin y(1 + \sin x)dy = 0 \).

Rearranging the terms, we get:
\( \frac{\cos x \, dx}{1 + \sin x} = \frac{\sin y \, dy}{1 + \cos y} \)

Integrating both sides, we get:
\( \int \frac{\cos x \, dx}{1 + \sin x} = \int \frac{\sin y \, dy}{1 + \cos y} + c \)

\( \log|1 + \sin x| = -\log|1 + \cos y| + \log c \)
\( \log|1 + \sin x| + \log|1 + \cos y| = \log c \)
\( (1 + \sin x)(1 + \cos y) = c \)

In simple words: Rearrange by moving cosine terms to one side and sine terms to the other. When you integrate cos(x)/(1 + sin(x)), use substitution: let u = 1 + sin(x), so du = cos(x)dx. This gives log|u|. Similarly for the right side with cosine. Then use log rules to get the product form.

Exam Tip: Watch for integrals of the form f'(x)/f(x), which always give log|f(x)|. Recognize that cos(x) is the derivative of sin(x), and -sin(y) is the derivative of cos(y), which makes these substitutions smooth.

Question 13. For each of the following differential equations, find a particular solution satisfying the given condition: \( \cos\left(\frac{dy}{dx}\right) = a \), where \( a \in \mathbb{R} \) and \( y = 2 \) when \( x = 0 \).
Answer: We have \( \cos\left(\frac{dy}{dx}\right) = a \).

\( \frac{dy}{dx} = \cos^{-1} a \)

\( dy = \cos^{-1} a \, dx \)

Integrating both sides, we get:
\( \int dy = \int \cos^{-1} a \, dx + c \)

\( y = x \cos^{-1} a + c \)

When \( x = 0, y = 2 \):
\( 2 = 0 + c \)
\( c = 2 \)

Therefore:
\( y = x \cos^{-1} a + 2 \)

Or equivalently:
\( \cos\left(\frac{y - 2}{x}\right) = a \)

In simple words: Since cosine of (dy/dx) equals a constant, take the inverse cosine of both sides to find dy/dx. Since (dy/dx) is now a constant, integration is straightforward. Substitute the initial condition x = 0, y = 2 to find the constant of integration.

Exam Tip: Always apply initial conditions after integration. When you have a constant on the right side of dy/dx, integrating gives a linear relationship between y and x. Check your answer by substituting the initial condition back.

Question 14. For each of the following differential equations, find a particular solution satisfying the given condition: \( \frac{dy}{dx} = -4xy^2 \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have \( \frac{dy}{dx} = -4xy^2 \).

Rearranging the terms, we get:
\( \frac{dy}{y^2} = -4x \, dx \)

Integrating both sides, we get:
\( \int \frac{dy}{y^2} = -\int 4x \, dx + c \)

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

When \( y = 1, x = 0 \):
\( (1)^{-1} = 2(0)^2 + c \)
\( 1 = c \)

Therefore:
\( \frac{1}{y} = 2x^2 + 1 \)
\( y = \frac{1}{2x^2 + 1} \)

In simple words: Separate variables by moving y² to the left and x to the right. Integrate both sides: 1/y² integrates to -1/y, and 4x integrates to 2x². Substitute the initial values to solve for the constant. Finally, rearrange to get y explicitly.

Exam Tip: When integrating negative powers like y^(-2), use the power rule: the antiderivative is y^(-1)/(-1) = -1/y. Always substitute initial conditions to find the constant before finalizing your answer.

Question 15. For each of the following differential equations, find a particular solution satisfying the given condition: \( x \, dy = (2x^2 + 1)dx \) (where \( x \neq 0 \)), given that \( y = 1 \) when \( x = 1 \).
Answer: We have \( x \, dy = (2x^2 + 1)dx \).

Rearranging the terms, we get:
\( dy = \frac{2x^2 + 1}{x}dx = \left(2x + \frac{1}{x}\right)dx \)

Integrating both sides, we get:
\( \int dy = \int 2x \, dx + \int \frac{1}{x}dx + c \)

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

When \( y = 1, x = 1 \):
\( 1 = 1^2 + \log 1 + c \)
\( 1 = 1 + 0 + c \)
\( c = 0 \)

Therefore:
\( y = x^2 + \log|x| \)

In simple words: Divide both sides by x to isolate dy. Split the resulting fraction into two simpler terms: 2x and 1/x. Integrate each piece separately - 2x gives x², and 1/x gives log|x|. Apply the initial condition to find the constant of integration, which turns out to be zero.

Exam Tip: When you have a fraction with multiple terms in the numerator, split it into separate fractions. Remember that the integral of 1/x is log|x|, not 1. Always verify your answer by checking the initial condition.

Question 16. For each of the following differential equations, find a particular solution satisfying the given condition: \( \frac{dy}{dx} = y \tan x \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have \( \frac{dy}{dx} = y \tan x \).

Rearranging the terms, we get:
\( \frac{dy}{y} = \tan x \, dx \)

Integrating both sides, we get:
\( \int \frac{dy}{y} = \int \tan x \, dx + c \)

\( \log|y| = \log|\sec x| + \log c \)
\( \log|y| - \log|\sec x| = \log c \)
\( \log\left|\frac{y}{\sec x}\right| = \log c \)
\( \frac{y}{\sec x} = c \)
\( y \cos x = c \)

When \( y = 1, x = 0 \):
\( 1 \times \cos 0 = c \)
\( c = 1 \)

Therefore:
\( y \cos x = 1 \)
\( y = \sec x \)

In simple words: Separate variables by moving all y terms to one side and all x terms to the other. The integral of 1/y is log|y|, and the integral of tan(x) is log|sec(x)| (or equivalently, -log|cos(x)|). Use logarithm properties to simplify. Then apply the initial condition to find c.

Exam Tip: Remember that \( \int \tan x \, dx = \log|\sec x| + C \) or equivalently \( -\log|\cos x| + C \). After simplifying with logarithm rules, verify your answer using the initial condition. In this case, the constant c = 1 makes the final answer particularly clean.

Exercise 19B

Question 1. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = \frac{x - 1}{y + 2} \)
Answer: We have \( \frac{dy}{dx} = \frac{x - 1}{y + 2} \).

\( (y + 2)dy = (x - 1)dx \)

Integrating on both sides:
\( \int (y + 2)dy = \int (x - 1)dx \)

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

\( y^2 + 4y - x^2 + 2x = C \)

In simple words: Cross-multiply to get all y terms on the left and all x terms on the right. Integrate each side separately. When you integrate (y + 2), you get y²/2 + 2y. When you integrate (x - 1), you get x²/2 - x. Multiply through by 2 to clear fractions and combine into a single equation.

Exam Tip: After separating variables, integrate each side carefully term-by-term. Multiply the equation by 2 at the end to avoid fractions, which makes the final answer cleaner and easier to verify.

Question 2. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = \frac{x}{x^2 + 1} \)
Answer: We have \( \frac{dy}{dx} = \frac{x}{x^2 + 1} \).

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

Multiply and divide 2 in numerator and denominator of the right side:
\( y = \frac{1}{2} \int \frac{2x}{x^2 + 1}dx \)

Integrating on both sides:
\( y = \frac{1}{2} \log(x^2 + 1) + C \)

In simple words: Notice that 2x is the derivative of x² + 1. To use this fact, multiply the numerator and denominator by 2. Now the numerator becomes the derivative of the denominator, so the integral is log(x² + 1). Don't forget to divide by 2 out front.

Exam Tip: Watch for integrals where the numerator is (or nearly is) the derivative of the denominator. Multiply or adjust to make it exact, then use the logarithm rule: \( \int \frac{f'(x)}{f(x)}dx = \log|f(x)| \).

Question 3. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} = (1 + x)(1 + y^2) \)
Answer: We have \( \frac{dy}{dx} = (1 + x)(1 + y^2) \).

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

Integrating on both sides:
\( \int \frac{dy}{1 + y^2} = \int (1 + x)dx \)

\( \tan^{-1} y = x + \frac{x^2}{2} + C \)

In simple words: Separate the variables by moving (1 + y²) to the left side as a denominator. Integrate both sides using standard formulas: the integral of 1/(1 + y²) is inverse tangent of y, and the integral of (1 + x) is x + x²/2.

Exam Tip: Remember that \( \int \frac{dy}{1 + y^2} = \tan^{-1} y \) - this is a standard formula that appears frequently. Recognize it immediately to speed up your work.

Question 4. Find the general solution of each of the following differential equations: \( (1 + x^2)\frac{dy}{dx} = xy \)
Answer: We have \( (1 + x^2)\frac{dy}{dx} = xy \).

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

Multiply and divide 2 in numerator and denominator of the right side:
\( \frac{1}{y}dy = \frac{1}{2} \cdot \frac{2x}{x^2 + 1}dx \)

Integrating on both sides:
\( \log y = \frac{1}{2} \log(1 + x^2) + \log C \)
\( \log y = \log\sqrt{1 + x^2} + \log C \)
\( y = \sqrt{1 + x^2} \cdot C_1 \)

In simple words: Separate variables so that y is on the left and x is on the right. On the right side, recognize that 2x is the derivative of (1 + x²), so multiply and divide by 2. Integrate both sides, then use log rules to simplify and solve for y.

Exam Tip: When you have \( \int \frac{dy}{y} \), the result is log y. The expression \( \frac{1}{2}\log(1 + x^2) \) can be rewritten as \( \log\sqrt{1 + x^2} \) using logarithm properties. Always simplify logarithmic answers fully.

Question 5. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} + y = 1(y \neq 1) \)
Answer: We have \( \frac{dy}{dx} + y = 1 \).

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

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

Integrating on both sides:
\( \int \frac{1}{1 - y}dy = \int dx \)

\( -\log|1 - y| = x + C \)
\( \log|1 - y| = -x + C' \)

In simple words: Rearrange to isolate dy/dx, then separate variables. When you integrate 1/(1 - y), remember that the derivative of (1 - y) is -1, so the integral gives -log|1 - y|. Keep the negative sign - it's crucial. You can rearrange further if needed, but this form shows the solution clearly.

Exam Tip: For integrals of the form \( \int \frac{dy}{1 - y} \), use substitution u = 1 - y, so du = -dy. This gives \( -\int \frac{du}{u} = -\log|u| = -\log|1 - y| \). Watch the negative sign carefully.

Question 6. Find the general solution of each of the following differential equations: \( \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0 \)
Answer: We have \( \frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0 \).

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

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

Integrating on both sides:
\( \int \frac{1}{\sqrt{1 - y^2}}dy = -\int \frac{1}{\sqrt{1 - x^2}}dx \)

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

In simple words: Rearrange and separate variables. Notice that \( \frac{1}{\sqrt{1 - y^2}} \) is the derivative of sin⁻¹(y), and \( \frac{1}{\sqrt{1 - x^2}} \) is the derivative of sin⁻¹(x). Integrate both sides using these standard formulas and rearrange to get the final form.

Exam Tip: Memorize the derivatives: d/dy[sin⁻¹(y)] = 1/√(1 - y²) and d/dx[sin⁻¹(x)] = 1/√(1 - x²). These appear frequently in integration problems. The final form with the sum equal to a constant is elegant and typical for this class of problems.

Question 7. Find the general solution of each of the following differential equations: \( x\frac{dy}{dx} + y = y^2 \)
Answer: We have \( x\frac{dy}{dx} + y = y^2 \).

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

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

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

Using partial fractions on the left side:
Let \( \frac{1}{y(y - 1)} = \frac{A}{y} + \frac{B}{y - 1} \)

\( 1 = A(y - 1) + By \)

Comparing coefficients:
\( A = -1, B = 1 \)

So: \( \frac{1}{y(y - 1)}dy = \left(-\frac{1}{y} + \frac{1}{y - 1}\right)dy \)

Integrating on both sides:
\( \int \left(-\frac{1}{y} + \frac{1}{y - 1}\right)dy = \int \frac{1}{x}dx \)

\( -\log y + \log(y - 1) = \log x + C \)
\( \log\left(\frac{y - 1}{y}\right) = \log x + C \)
\( \frac{y - 1}{y} = x \cdot C_1 \)
\( y - 1 = xy \cdot C_1 \)
\( y(1 - xC_1) = 1 \)
\( y = \frac{1}{1 - xC} \)

In simple words: Rearrange to separate variables. The left side has a product y(y - 1) in the denominator, so use partial fractions to break it into two simpler fractions, each with a single linear factor. Find constants A and B by comparing coefficients. Then integrate each fraction separately and combine using log rules.

Exam Tip: Partial fractions is essential for rational integrands. Set up the decomposition correctly, substitute convenient values to find coefficients, and integrate each piece. After integration, use log properties to combine logarithms and simplify the final form.

Question 8. Find the general solution of each of the following differential equations: \( x^2(y + 1)dx + y^2(x - 1)dy = 0 \)
Answer: We have \( x^2(y + 1)dx + y^2(x - 1)dy = 0 \).

\( x^2(y + 1)dx = -y^2(x - 1)dy \)
\( x^2(y + 1)dx = y^2(1 - x)dy \)

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

Add and subtract 1 in numerators of both sides:
\( \frac{x^2 - 1 + 1}{1 - x}dx = \frac{y^2 - 1 + 1}{y + 1}dy \)

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

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

Splitting the terms:
\( \left[-(x + 1) + \frac{1}{1 - x}\right]dx = \left[(y - 1) + \frac{1}{y + 1}\right]dy \)

Integrating:
\( \int \left[-(x + 1) + \frac{1}{1 - x}\right]dx = \int \left[(y - 1) + \frac{1}{y + 1}\right]dy \)

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

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

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

In simple words: Separate the variables by rearranging the equation. To integrate the resulting fractions, use polynomial long division or rewrite by adding and subtracting constants in the numerator. This breaks each fraction into simpler parts - a polynomial and a simple fraction. Integrate each piece separately, then combine all logarithmic and polynomial terms.

Exam Tip: When you have a rational integrand where the numerator's degree is greater than or equal to the denominator's degree, use polynomial long division or add-and-subtract technique first. This breaks it into easier parts: a polynomial plus a proper fraction. Always expand (a - b)(a + b) = a² - b² to help with the algebra.

Question 9. Find the general solution of each of the following differential equations: \( y(1 - x^2)\frac{dy}{dx} = x(1 + y^2) \)
Answer: We have \( y(1 - x^2)\frac{dy}{dx} = x(1 + y^2) \).

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

Multiply 2 in both sides:
\( \frac{2y}{1 + y^2}dy = \frac{2x}{1 - x^2}dx \)

Integrating on both sides:
\( \int \frac{2y}{1 + y^2}dy = \int \frac{2x}{1 - x^2}dx \)

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

In simple words: Separate variables carefully. Multiply both sides by 2 to create numerators that are derivatives of their respective denominators: 2y is the derivative of (1 + y²), and 2x is the derivative of (1 - x²). Use the log rule \( \int \frac{f'(u)}{f(u)}du = \log|f(u)| \) on both sides. Combine logs and exponentiate to get the final form.

Exam Tip: Watch for the pattern where the numerator is (or almost is) the derivative of the denominator. Multiplying by 2 is a standard trick to make this exact. After integration, combine logarithms using log(a) + log(b) = log(ab) to get the product form.

Question 10. Find the general solution of each of the following differential equations: \( y \log y \, dx - x \, dy = 0 \)
Answer: We have \( y \log y \, dx - x \, dy = 0 \).

\( y \log y \, dx = x \, dy \)

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

Integrating on both sides:
\( \int \frac{1}{x}dx = \int \frac{1}{y \log y}dy \)

For the left side:
\( \log x \)

For the right side, let \( t = \log y \), so \( \frac{1}{y}dy = dt \):
\( \int \frac{1}{y \log y}dy = \int \frac{1}{t}dt = \log t = \log(\log y) \)

Therefore:
\( \log x = \log(\log y) + \log C \)
\( x = (\log y) \cdot C \)

In simple words: Rearrange to separate variables. The right side requires substitution: let t = log(y), so dt = (1/y)dy. The integral becomes 1/t, which is log(t) = log(log(y)). After integrating both sides, use log properties to solve for x.

Exam Tip: When you encounter nested functions like log(log y), use substitution to simplify. Setting u = log(y) turns a complex integral into a standard one. Always write the final answer by exponentiation or logarithm simplification to match the expected form.

Question 11. Find the general solution of each of the following differential equations: \( x(x^2 - x^2 y^2)dy + y(y^2 + x^2 y^2)dx = 0 \)
Answer: We have \( x(x^2 - x^2 y^2)dy + y(y^2 + x^2 y^2)dx = 0 \).

\( x \cdot x^2(1 - y^2)dy + y \cdot y^2(1 + x^2)dx = 0 \)
\( x^3(1 - y^2)dy + y^3(1 + x^2)dx = 0 \)

\( \frac{1 + x^2}{x^3}dx + \frac{1 - y^2}{y^3}dy = 0 \)

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

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

Integrating:
\( \int \frac{1}{x^3}dx + \int \frac{1}{x}dx = -\int \frac{1}{y^3}dy + \int \frac{1}{y}dy + C \)

\( \frac{x^{-3 + 1}}{-3 + 1} + \log x = -\frac{y^{-3 + 1}}{-3 + 1} + \log y + C \)

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

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

\( -\frac{1}{2x^2} - \frac{1}{2y^2} + \log\left(\frac{x}{y}\right) = C \)

In simple words: Start by factoring out common terms: the first part factors as x³(1 - y²) and the second as y³(1 + x²). Rearrange to separate variables, then split each fraction into separate terms. Integrate using the power rule (for negative powers) and the logarithm rule. Combine like terms and use log properties.

Exam Tip: Factor completely before separating variables - this often reveals a much simpler form. When you split a fraction like (1 + x²)/x³ into 1/x³ + 1/x, each piece integrates easily. Remember that \( \int x^{-3}dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} \).

Question 12. Find the general solution of each of the following differential equations: \( (1 - x^2)dy + xy(1 - y)dx = 0 \)
Answer: We have \( (1 - x^2)dy + xy(1 - y)dx = 0 \).

\( (1 - x^2)dy = -xy(1 - y)dx \)
\( (1 - x^2)dy = xy(y - 1)dx \)

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

Integrating on both sides:
\( \int \frac{1}{y(y - 1)}dy = \int \frac{x}{1 - x^2}dx \)

For the left side, use partial fractions:
Let \( \frac{1}{y(y - 1)} = \frac{A}{y} + \frac{B}{y - 1} \)
\( 1 = A(y - 1) + By \)
\( A = -1, B = 1 \)

\( \int \left(-\frac{1}{y} + \frac{1}{y - 1}\right)dy = -\log y + \log(y - 1) = \log\left(\frac{y - 1}{y}\right) \)

For the right side, multiply and divide 2:
\( \int \frac{x}{1 - x^2}dx = \frac{1}{2}\int \frac{2x}{1 - x^2}dx = -\frac{1}{2}\log(1 - x^2) = \log\sqrt{\frac{1}{1 - x^2}} + \log C \)

Therefore:
\( \log\left(\frac{y - 1}{y}\right) = -\log\sqrt{1 - x^2} + \log C \)
\( \log\left(\frac{y - 1}{y}\right) = \log\sqrt{1 - x^2} + \log C \)
\( \frac{y - 1}{y} = \sqrt{1 - x^2} \cdot C \)
\( y = (y - 1)\sqrt{1 - x^2} \cdot C \)
\( y = \sqrt{1 - x^2} \cdot C \)

In simple words: Rearrange to isolate dy on one side. Use partial fractions to decompose 1/[y(y - 1)] into two simple fractions. On the right, recognize that 2x is the derivative of (1 - x²), so multiply and divide by 2. Integrate both sides, combine logarithms, and exponentiate to get the final form with the product of a square root and a constant.

Exam Tip: Partial fractions and the derivative-in-numerator pattern (d/dx[1 - x²] = -2x) are key to solving this problem efficiently. After exponentiation, simplify carefully - the square root in the exponent becomes a square root in the coefficient.

Question 1. Find the general solution of the differential equation: (1 - x²)(1 - y) dx = xy (1 + y) dy
Answer: Rearranging the equation, we get \( \frac{1 - x^2}{x} \, dx = \frac{y(1 + y)}{(1 - y)} \, dy \). Breaking down the left side: \( \left[\frac{1}{x} - x\right] dx = \left[\frac{y + y^2}{1 - y}\right] dy \). For the right side, decompose by adding and subtracting 1 in the numerator: \( \frac{y^2 - 1 + 1}{(1 - y)} dy = \frac{(y - 1)(y + 1) + 1}{(1 - y)} dy = \left[\frac{y + y^2}{1 - y}\right] dy = \left[\frac{y}{1 - y} + \frac{y^2}{1 - y}\right] dy \). Integrating both sides: \( \int \left[\frac{1}{x} - x\right] dx = \int \left[\frac{y}{1 - y} + \frac{y^2}{1 - y}\right] dy \). Left side gives: \( \log x - \frac{x^2}{2} \). For the right side, decompose the first term: \( \int \frac{y}{1 - y} dy = \int \frac{y - 1 + 1}{1 - y} dy = \int (-1) dy + \int \frac{1}{1 - y} dy = -y + \log|1 - y| \). For the second term, use the identity \( (a^2 - b^2) = (a + b)(a - b) \): \( \frac{(y + 1)(y - 1) + 1}{(1 - y)} dy = -\frac{(y + 1)(y - 1) + 1}{(y - 1)} dy = -(y + 1) dy + \frac{1}{(1 - y)} dy \). Integrating: \( \int -(y + 1) dy - \int \frac{1}{(y - 1)} dy = -\frac{y^2}{2} - y - \log|y - 1| \). Combining all results, the general solution is: \( \log x - \frac{x^2}{2} = -y + \log|1 - y| - \frac{y^2}{2} - y - \log|y - 1| + C \), which simplifies to \( \log x \cdot \frac{x^2}{2} = -y + \log|1 - y| - \left(\frac{y^2}{2} + y\right) + \log|y - 1| + C \), or in cleaner form: \( \log[x.(1 - y)^2] = \frac{x^2}{2} - \frac{y^2}{2} - 2y + C \).

Exam Tip: When handling rational expressions with polynomial numerators and denominators, always use polynomial division or decomposition (add and subtract terms) to break down complex fractions into simpler, integrable parts.

Question 2. Find the general solution of the differential equation: (y + xy) dx + (x - xy²) dy = 0
Answer: Factor out common terms: \( y(1 + x) dx + x(1 - y^2) dy = 0 \). Rearrange to separate variables: \( \frac{1 + x}{x} dx + \frac{1 - y^2}{y} dy = 0 \). Expand the fractions: \( \left[\frac{1}{x} + 1\right] dx + \left[\frac{1}{y} - y\right] dy = 0 \). Integrate both sides: \( \int \frac{1}{x} dx + \int 1 \, dx + \int \frac{1}{y} dy - \int y \, dy = C \). This yields: \( \log|x| + x + \log|y| - \frac{y^2}{2} = C \), or equivalently: \( \log|xy| + x - \frac{y^2}{2} = C \).

Exam Tip: Always factor expressions before attempting separation of variables; this step often reveals the natural structure of the equation and simplifies the integration process significantly.

Question 3. Find the general solution of the differential equation: (x² - yx²) dy + (y² + xy²) dx = 0
Answer: Extract common factors: \( x^2(1 - y) dy + y^2(1 + x) dx = 0 \). Rearrange: \( \frac{1 + x}{y^2} dx + \frac{1 - y}{x^2} dy = 0 \). Separate the fractions: \( \left[\frac{1}{y^2} + \frac{x}{y^2}\right] dx + \left[\frac{1}{x^2} - \frac{y}{x^2}\right] dy = 0 \). Rewrite: \( \frac{1}{y^2} dx + \frac{1}{x} dx + \frac{1}{x^2} dy - \frac{1}{y} dy = 0 \). Integrate: \( \int \frac{1}{y^2} dx + \int \frac{1}{x} dx + \int \frac{1}{x^2} dy - \int \frac{1}{y} dy = C \). Evaluating: \( -\frac{1}{x} + \log|x| + \frac{1}{y} - \log|y| = C \), which can be rewritten as: \( \log\left|\frac{x}{y}\right| = \frac{1}{x} + \frac{1}{y} + C \).

Exam Tip: After factoring, organize terms so that each integral operates on only one variable at a time; this discipline prevents errors and clarifies the structure of the final answer.

Question 4. Find the general solution of the differential equation: (x²y - x²) dx + (xy² - y²) dy = 0
Answer: Pull out common factors: \( x^2(y - 1) dx + y^2(x - 1) dy = 0 \). Rearrange: \( \frac{x^2}{y^2} dx + \frac{y^2}{x^2} dy = 0 \) after dividing by \( (y - 1)(x - 1) \). Decompose by adding and subtracting 1 in the numerators. For the x-terms: \( \frac{x^2 - 1 + 1}{(x - 1)} dx \), which expands to \( \frac{(x + 1)(x - 1) + 1}{(x - 1)} dx = (x + 1) dx + \frac{1}{(x - 1)} dx \). Similarly for y-terms: \( (y + 1) dy + \frac{1}{(y - 1)} dy \). Integrate all terms: \( \int (x + 1) dx + \int \frac{1}{(x - 1)} dx + \int (y + 1) dy + \int \frac{1}{(y - 1)} dy = C \). This produces: \( \frac{x^2}{2} + x + \log|x - 1| + \frac{y^2}{2} + y + \log|y - 1| = C \), which simplifies to: \( \frac{1}{2}(x^2 + y^2) + (x + y) + \log|(x - 1)(y - 1)| = C \).

Exam Tip: When you see a quadratic numerator and linear denominator, always think about the algebraic identity \( a^2 - b^2 = (a + b)(a - b) \); this often unlocks the decomposition needed for integration.

Question 5. Find the general solution of the differential equation: x\(\sqrt{1 + y^2}\) dx + y\(\sqrt{1 + x^2}\) dy = 0
Answer: Rearrange to isolate square root terms: \( \frac{x}{\sqrt{1 + x^2}} dx + \frac{y}{\sqrt{1 + y^2}} dy = 0 \). Recognize that \( \frac{d}{dx}\sqrt{1 + x^2} = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}} \), which means each integral evaluates directly. Integrating both sides: \( \int \frac{x}{\sqrt{1 + x^2}} dx + \int \frac{y}{\sqrt{1 + y^2}} dy = C \). Both integrals yield square root expressions: \( \sqrt{1 + x^2} + \sqrt{1 + y^2} = C \).

Exam Tip: Always examine the integrand structure to see if it matches a known derivative formula; recognizing that \( \frac{x}{\sqrt{1 + x^2}} \) is the derivative of \( \sqrt{1 + x^2} \) saves time and reduces calculation errors.

Question 6. Find the general solution of the differential equation: \( \frac{dy}{dx} = e^{x + y} + x^2 e^y \)
Answer: Expand the right side: \( \frac{dy}{dx} = e^x \cdot e^y + x^2 e^y \). Factor: \( \frac{dy}{dx} = e^y(e^x + x^2) \). Separate variables: \( \frac{1}{e^y} dy = (e^x + x^2) dx \), or equivalently \( e^{-y} dy = (e^x + x^2) dx \). Integrate both sides: \( \int e^{-y} dy = \int e^x dx + \int x^2 dx \). Evaluating each integral: \( -e^{-y} = e^x + \frac{x^3}{3} + C \). Rearranging: \( e^x + e^{-y} + \frac{x^3}{3} = C \).

Exam Tip: When the exponent in an exponential expression contains a sum like \( e^{x+y} \), always split it as a product of separate exponentials; this immediately reveals opportunities for variable separation.

Question 7. Find the general solution of the differential equation: \( \frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}} \)
Answer: Simplify the denominator by rewriting \( e^x + e^{-x} \) with a common factor: \( e^x + e^{-x} = e^x + \frac{1}{e^x} = \frac{e^{2x} + 1}{e^x} \). Substitute this into the right side: \( \frac{dy}{dx} = \frac{(3e^{2x} + 3e^{4x}) \cdot e^x}{e^{2x} + 1} = \frac{3e^{3x}(1 + e^{2x}) \cdot e^x}{e^{2x} + 1} = 3e^{3x} \). This simplifies to: \( dy = 3e^{3x} dx \). Integrate both sides: \( y = \int 3e^{3x} dx = 3 \cdot \frac{e^{3x}}{3} + C = e^{3x} + C \).

Exam Tip: Always try to simplify complex rational expressions by factoring out common terms or combining denominators; this often reveals hidden cancellations that dramatically reduce calculation effort.

Question 8. Find the general solution of the differential equation: (\(\cos x\))\(\frac{dy}{dx}\) + \(\cos 2x\) = \(\cos 3x\)
Answer: Rearrange: \( \frac{dy}{dx} = \frac{\cos 3x - \cos 2x}{\cos x} \). Use trigonometric sum-to-product identities to simplify the numerator. Recall that \( \cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \). Applying this: \( \cos 3x - \cos 2x = -2\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right) \). Alternatively, work through the original approach: multiply numerator and denominator by appropriate factors to reveal integrable forms. After algebraic manipulation (expanding products and using double-angle formulas), the expression simplifies through cancellation to: \( \frac{dy}{dx} = \cos 2x - 2\sin^2 x - 2\cos x + \sec x \). Integrate each term: \( y = \int \cos 2x \, dx - \int 2\sin^2 x \, dx - \int 2\cos x \, dx + \int \sec x \, dx \). Evaluating: \( y = \frac{\sin 2x}{2} - \int(1 - \cos 2x) dx - 2\sin x + \log|\sec x + \tan x| + C = \frac{\sin 2x}{2} - 2\sin x - x + \log|\sec x + \tan x| + C \).

Exam Tip: For trigonometric differential equations, always recall the standard product-to-sum and sum-to-product identities; they often transform unwieldy expressions into manageable forms that integrate directly.

Question 9. Find the general solution of the differential equation: 3e\(^x\) \(\tan y\) dx + (1 - e\(^x\))\(\sec^2 y\) dy = 0
Answer: Rearrange: \( 3e^x \tan y \, dx = (e^x - 1) \sec^2 y \, dy \). Separate variables: \( 3 \cdot \frac{e^x}{e^x - 1} dx = \frac{\sec^2 y}{\tan y} dy \). For the left side, use substitution or recognize the derivative pattern: the numerator is almost the derivative of the denominator. Rewrite: \( 3 \cdot \frac{1}{e^x - 1} dx = 3 \cdot \frac{1}{1 - e^{-x}} \cdot e^{-x} dx \). Integrate: \( 3\log|1 - e^{-x}| \) (or equivalently \( 3\log|e^x - 1| - 3x \)). For the right side: \( \int \frac{\sec^2 y}{\tan y} dy = \int \frac{1}{\sin y \cos y} dy \). Let \( u = \sin y \), so \( du = \cos y \, dy \), and this integral becomes \( \log|\tan y| \). Combining: \( 3\log|1 - e^{-x}| = \log|\tan y| + \log C \), which simplifies to: \( (1 - e^{-x})^3 = \tan y \cdot C \) or \( \tan y = (1 - e^{-x})^3 \cdot C \).

Exam Tip: When dealing with exponential terms in the denominator, always consider whether a substitution or rewriting will expose a standard logarithmic derivative; this technique is especially powerful for integrals of the form \( \int \frac{1}{a - be^x} dx \).

Question 10. Find the general solution of the differential equation: \( \frac{dy}{dx} + \sin(x + y) = \sin(x - y) \)
Answer: Use trigonometric difference formulas: \( \sin(x + y) = \sin x \cos y + \cos x \sin y \) and \( \sin(x - y) = \sin x \cos y - \cos x \sin y \). Substitute: \( \frac{dy}{dx} + \sin x \cos y + \cos x \sin y = \sin x \cos y - \cos x \sin y \). Simplify by canceling \( \sin x \cos y \): \( \frac{dy}{dx} = -2\cos x \sin y \), or equivalently \( \frac{dy}{dx} = -2\sin y \cos x \). Separate variables: \( \frac{dy}{\sin y} = -2\cos x \, dx \), which becomes \( \csc y \, dy = -2\cos x \, dx \). Integrate both sides: \( -\log|\csc y + \cot y| = -2\sin x + C \). This gives: \( \log|\csc y - \cot y| = 2\sin x + C \), or in simplified form: \( \sin x + \log|\csc y - \cot y| + C = 0 \).

Exam Tip: Always expand sums and differences of angles using angle-addition formulas first; this reveals cancellations and variable-separation opportunities that are not apparent from the factored form.

Question 11. Find the general solution of the differential equation: e\(^y\)(1 + x²) dy - \(\frac{x}{y}\) dx = 0
Answer: Rearrange: \( e^y(1 + x^2) dy = \frac{x}{y} dx \), which gives \( e^y \cdot y \, dy = \frac{x}{1 + x^2} dx \). Integrate both sides: \( \int y e^y dy = \int \frac{x}{1 + x^2} dx \). For the left side, apply integration by parts with \( u = y \) and \( dv = e^y dy \): \( \int y e^y dy = y e^y - \int e^y dy = y e^y - e^y = e^y(y - 1) \). For the right side, use substitution \( w = 1 + x^2 \), so \( dw = 2x \, dx \): \( \int \frac{x}{1 + x^2} dx = \frac{1}{2}\log(1 + x^2) \). Combining: \( e^y(y - 1) = \frac{1}{2}\log(1 + x^2) + C \).

Exam Tip: When an integral involves a product like \( y e^y \), immediately recognize that integration by parts is needed; the key is to choose \( u \) as the polynomial term so its derivative simplifies the remaining integral.

Question 12. Find the general solution of the differential equation: \( \frac{dy}{dx} + \frac{(1 + \cos 2y)}{(1 - \cos 2x)} = 0 \)
Answer: Rearrange: \( \frac{dy}{dx} = -\frac{1 + \cos 2y}{1 - \cos 2x} \). Apply double-angle identities: \( 1 + \cos 2y = 2\cos^2 y \) and \( 1 - \cos 2x = 2\sin^2 x \). Substitute: \( \frac{dy}{dx} = -\frac{2\cos^2 y}{2\sin^2 x} = -\frac{\cos^2 y}{\sin^2 x} \). Separate variables: \( \sec^2 y \, dy = -\csc^2 x \, dx \). Integrate both sides: \( \int \sec^2 y \, dy = -\int \csc^2 x \, dx \). Evaluating: \( \tan y = \cot x + C \).

Exam Tip: Always convert double-angle expressions like \( 1 \pm \cos 2\theta \) to squared single-angle forms; this often reveals hidden trigonometric functions that separate cleanly.

Question 13. Find the general solution of the differential equation: \( \frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0 \)
Answer: Rearrange: \( \frac{dy}{dx} = -\cos x \tan y \). Separate variables: \( \frac{dy}{\tan y} = -\cos x \, dx \), which becomes \( \cot y \, dy = -\cos x \, dx \). Integrate both sides: \( \int \cot y \, dy = -\int \cos x \, dx \). Evaluating: \( \log|\sin y| = -\sin x + C \), which can be rewritten as: \( \sin y = e^{-\sin x + C} = A e^{-\sin x} \) where \( A = e^C \).

Exam Tip: Trigonometric cotangent and tangent integrals appear frequently; remember that \( \int \cot \theta \, d\theta = \log|\sin \theta| + C \) and commit this to memory to speed up calculations.

Question 14. Find the general solution of the differential equation: \(\cos x\)(1 + \(\cos y\))dx - \(\sin y\)(1 + \(\sin x\))dy = 0
Answer: Divide the entire equation by (1 + \(\sin x\))(1 + \(\cos y\)): \( \frac{\cos x \, dx}{1 + \sin x} - \frac{\sin y \, dy}{1 + \cos y} = 0 \). Recognize that \( \frac{d}{dx}(1 + \sin x) = \cos x \) and \( \frac{d}{dy}(1 + \cos y) = -\sin y \). This means: \( \frac{d(1 + \sin x)}{1 + \sin x} = \frac{\sin y \, dy}{1 + \cos y} \). Integrate both sides: \( \log|1 + \sin x| = -\log|1 + \cos y| + \log C \). Simplifying: \( \log|1 + \sin x| + \log|1 + \cos y| = \log C \), which gives: \( (1 + \sin x)(1 + \cos y) = C \).

Exam Tip: When encountering equations with products in the numerator and denominator, always attempt division by a strategic product that will create exact derivatives in the numerators; this is a hallmark of equations that integrate to logarithmic forms.

Question 15. Find the general solution of the differential equation: \( \frac{dy}{dx} = e^{x + y} + e^{x - y} \)
Answer: Expand: \( \frac{dy}{dx} = e^x e^y + e^x e^{-y} = e^x(e^y + e^{-y}) \). Separate variables: \( \frac{dy}{e^y + e^{-y}} = e^x dx \). The denominator can be rewritten: \( e^y + e^{-y} = \frac{e^{2y} + 1}{e^y} \), so \( \frac{e^y}{e^{2y} + 1} dy = e^x dx \). Integrate the left side by letting \( u = e^y \), so \( du = e^y dy \): \( \int \frac{du}{u^2 + 1} = \tan^{-1}(u) + \text{const} = \tan^{-1}(e^y) \). Integrate the right side: \( \int e^x dx = e^x \). Combining: \( \tan^{-1}(e^y) = e^x + C \), or equivalently: \( e^y = \tan(e^x + C) \).

Exam Tip: When separating variables produces a denominator of the form \( e^y + e^{-y} \), multiply both numerator and denominator by \( e^y \) to create a rational expression in \( e^y \), which then integrates using standard arctangent substitution.

Question 16. Find the general solution of the differential equation: \( \sin^3 x \, dx - \sin y \, dy = 0 \)
Answer: Apply the identity \( \sin^3 x = \frac{3\sin x - \sin 3x}{4} \). Substitute: \( \frac{3\sin x - \sin 3x}{4} dx = \sin y \, dy \). Integrate both sides: \( \int \frac{3\sin x - \sin 3x}{4} dx = \int \sin y \, dy \). For the left side: \( \frac{1}{4}\left(-3\cos x + \frac{\cos 3x}{3}\right) = \int \sin y \, dy \). Evaluating the right side: \( -\cos y \). Combining all terms: \( \frac{1}{4}\left(-3\cos x + \frac{\cos 3x}{3}\right) = -\cos y + C \), which simplifies to: \( 12\cos y + \cos 3x - 9\cos x = C \).

Exam Tip: The cubic sine formula \( \sin^3 \theta = \frac{3\sin \theta - \sin 3\theta}{4} \) is essential for integrating odd powers of sine; memorize it to avoid deriving it during an exam.

Question 17. Find the general solution of the differential equation: (e\(^y\) + 1)\(\cos x\) dx + e\(^y\) \(\sin x\) dy = 0
Answer: Rearrange: \( (e^y + 1)\cos x \, dx = -e^y \sin x \, dy \). Separate: \( \frac{\cos x}{\sin x} dx = -\frac{e^y}{e^y + 1} dy \), which becomes \( \cot x \, dx = -\frac{e^y}{e^y + 1} dy \). Integrate both sides: \( \int \cot x \, dx = -\int \frac{e^y}{e^y + 1} dy \). The left integral gives \( \log|\sin x| \). For the right integral, let \( u = e^y + 1 \), so \( du = e^y dy \): \( -\int \frac{du}{u} = -\log|e^y + 1| \). Combining: \( \log|\sin x| + \log|e^y + 1| = \log C \), which simplifies to: \( \sin x \cdot (e^y + 1) = C \).

Exam Tip: When a fraction has the same function in both numerator and denominator (like \( \frac{e^y}{e^y + 1} \)), always consider substitution \( u = \text{denominator} \); this converts the integral into the standard logarithmic form \( \int \frac{du}{u} \).

Question 18. Find the general solution of the differential equation: \( \frac{dy}{dx} + \sin(x + y) = \sin(x - y) \)
Answer: Expand using angle formulas: \( \frac{dy}{dx} + \sin x \cos y + \cos x \sin y = \sin x \cos y - \cos x \sin y \). Cancel \( \sin x \cos y \) terms: \( \frac{dy}{dx} = -2\cos x \sin y \). Separate variables: \( \csc y \, dy = -2\cos x \, dx \). Integrate: \( -\log|\csc y + \cot y| = -2\sin x + C \). Rearranging: \( \log|\csc y - \cot y| = 2\sin x + C \), or in combined form: \( \sin x + \log|\csc y - \cot y| + C = 0 \).

Exam Tip: The integral \( \int \csc y \, dy = -\log|\csc y + \cot y| + C \) (or equivalently \( \log|\csc y - \cot y| + C \)) should be memorized; it appears frequently in trigonometric differential equations.

Question 19. Find the general solution of the differential equation: \( \frac{dy}{dx} = \frac{xy + y}{xy + x} \)
Answer: Factor numerator and denominator: \( \frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)} \). Separate variables: \( \frac{y}{y + 1} dy = \frac{x + 1}{x} dx \). Expand the fractions: \( \left[1 - \frac{1}{y + 1}\right] dy = \left[1 + \frac{1}{x}\right] dx \). Integrate both sides: \( y - \log|y + 1| = x + \log|x| + C \). Rearranging: \( \log|xy| + x + y = C \).

Exam Tip: When integrating rational expressions like \( \frac{y}{y + 1} \), always rewrite by polynomial division to isolate the improper fraction part; this reveals simpler integrands.

Question 20. Find the general solution of the differential equation: \( \frac{1}{x} \cos^2 y \, dy + \frac{1}{y} \cos^2 x \, dx = 0 \)
Answer: Rearrange: \( \frac{\cos^2 x}{x} dx + \frac{\cos^2 y}{y} dy = 0 \). Use the identity \( 2\cos^2 \theta = 1 + \cos 2\theta \): \( \frac{1 + \cos 2x}{2x} dx + \frac{1 + \cos 2y}{2y} dy = 0 \). Integrate both sides: \( \frac{1}{2}\int \frac{1}{x} dx + \frac{1}{2}\int \frac{\cos 2x}{x} dx + \frac{1}{2}\int \frac{1}{y} dy + \frac{1}{2}\int \frac{\cos 2y}{y} dy = C \). Evaluating: \( \frac{1}{2}\log|x| + \frac{1}{2}\log|y| + \frac{1}{4}(\sin 2x + \sin 2y) = C \). This simplifies to: \( \log|xy| + \sin 2x + \sin 2y = C \). Or in a cleaner form: \( \frac{x^2}{2} + \frac{y^2}{2} + \sin 2x + \sin 2y = C \).

Exam Tip: Always apply the double-angle identity \( 2\cos^2 \theta = 1 + \cos 2\theta \) when dealing with squared trigonometric functions; it transforms integrals from challenging to straightforward.

Question 21. Find the particular solution of the differential equation \( \frac{dy}{dx} = 1 + x + y + xy \), given that y = 0 when x = 1.
Answer: Factor the right side: \( \frac{dy}{dx} = (1 + x)(1 + y) \). Separate variables: \( \frac{dy}{1 + y} = (1 + x) dx \). Integrate both sides: \( \log|1 + y| = x + \frac{x^2}{2} + C \). Apply the initial condition \( y = 0 \) when \( x = 1 \): \( \log(1) = 1 + \frac{1}{2} + C \), so \( 0 = \frac{3}{2} + C \), giving \( C = -\frac{3}{2} \). Substitute back: \( \log|1 + y| = x + \frac{x^2}{2} - \frac{3}{2} \). Exponentiating: \( 1 + y = e^{x + \frac{x^2}{2} - \frac{3}{2}} \), or \( y = e^{x + \frac{x^2}{2} - \frac{3}{2}} - 1 \).

Exam Tip: Always substitute the initial condition into the general solution immediately after integration to find the constant of integration; this step ensures the particular solution satisfies the given constraint.

Question 22. Find the particular solution of the differential equation x(1 + y²) dx - y(1 + x²) dy = 0, given that y = 1 when x = 0.
Answer: Separate variables: \( \frac{x}{1 + x^2} dx = \frac{y}{1 + y^2} dy \). Integrate both sides: \( \int \frac{x}{1 + x^2} dx = \int \frac{y}{1 + y^2} dy \). Using substitution (let \( u = 1 + x^2 \) and \( v = 1 + y^2 \)): \( \frac{1}{2}\log(1 + x^2) = \frac{1}{2}\log(1 + y^2) + C \), which simplifies to: \( \log(1 + x^2) = \log(1 + y^2) + \log K \), or \( (1 + x^2) = K(1 + y^2) \). Apply the initial condition \( y = 1 \) when \( x = 0 \): \( 1 = K(2) \), so \( K = \frac{1}{2} \). The particular solution is: \( (1 + x^2) = \frac{1}{2}(1 + y^2) \), which gives \( 2(1 + x^2) = 1 + y^2 \), or \( y^2 = 2x^2 + 1 \), so \( y = \sqrt{2x^2 + 1} \).

Exam Tip: For particular solutions, always substitute the initial condition into the general solution before simplifying; failure to do this leads to incorrect final answers.

Question 23. Find the general solution of the differential equation: e\(^x\) \(\sqrt{1 - y^2}\) dx + \(\frac{y}{x}\) dy = 0
Answer: Rearrange: \( e^x \sqrt{1 - y^2} \, dx = -\frac{y}{x} dy \). Separate: \( x e^x dx = -\frac{y}{\sqrt{1 - y^2}} dy \). Integrate both sides: \( \int x e^x dx = -\int \frac{y}{\sqrt{1 - y^2}} dy \). For the left integral, use integration by parts with \( u = x \) and \( dv = e^x dx \): \( x e^x - \int e^x dx = x e^x - e^x = e^x(x - 1) \). For the right integral, substitute \( w = 1 - y^2 \), so \( dw = -2y \, dy \): \( -\int \frac{-dw/2}{\sqrt{w}} = \int \frac{dw}{2\sqrt{w}} = \sqrt{w} = \sqrt{1 - y^2} \). Combining: \( e^x(x - 1) = \sqrt{1 - y^2} + C \), or \( e^x(x - 1) - \sqrt{1 - y^2} = C \).

Exam Tip: Always recognize when an integral involves a product of an algebraic function and an exponential; integration by parts with the polynomial as \( u \) and the exponential as \( dv \) is the standard approach.

Question 24. Find the general solution of the differential equation: \( \cos x \, \log y \, \frac{dy}{dx} + x^2 y = 0 \)
Answer: Rearrange: \( \cos x \, \log y \, dy = -x^2 y \, dx \), or \( \frac{\log y}{y} dy = -\frac{x^2}{\cos x} dx \). Rewrite the right side: \( \frac{\log y}{y} dy = -x^2 \sec x \, dx \). Integrate both sides: \( \int \frac{\log y}{y} dy = -\int x^2 \sec x \, dx \). For the left integral, let \( t = \log y \), so \( dt = \frac{dy}{y} \): \( \int t \, dt = \frac{(\log y)^2}{2} \). For the right integral, use integration by parts twice (or recognize it as a tabulated form): \( -\int x^2 \sec x \, dx = -x^2 \tan x + 2x \sec x - 2\log|\sec x + \tan x| \). Combining: \( \frac{(\log y)^2}{2} = -x^2 \tan x + 2x \sec x - 2\log|\sec x + \tan x| + C \).

Exam Tip: When integrating products of algebraic and trigonometric functions like \( x^2 \sec x \), successive integration by parts is required; this technique is computationally intensive but systematic — choose algebraic as \( u \) each time.

Question 25. Find the general solution of the differential equation: \( \sqrt{1 - x^4} \, dy = x \, dx \)
Answer: Separate variables: \( dy = \frac{x}{\sqrt{1 - x^4}} dx \). Multiply and divide by 2: \( dy = \frac{1}{2} \cdot \frac{2x}{\sqrt{1 - x^4}} dx \). Recognize that \( 1 - x^4 = 1 - (x^2)^2 \), so \( \frac{d}{dx}[\sin^{-1}(x^2)] = \frac{2x}{\sqrt{1 - x^4}} \). Integrate: \( \int dy = \frac{1}{2} \int \frac{2x}{\sqrt{1 - x^4}} dx = \frac{1}{2} \sin^{-1}(x^2) + C \). The solution is: \( y = \frac{1}{2} \sin^{-1}(x^2) + C \).

Exam Tip: When you encounter an expression like \( \frac{1}{\sqrt{1 - (u)^2}} \), immediately think inverse sine; matching the derivative formula \( \frac{d}{dx}[\sin^{-1}(u)] = \frac{u'}{\sqrt{1 - u^2}} \) saves time and prevents errors.

Question 26. Find the general solution of the differential equation: \( \frac{dy}{dx} = \sin^3 x \cos^2 x + xe^x \)
Answer: This is a direct integration equation: \( y = \int (\sin^3 x \cos^2 x + xe^x) dx \). For the first integral, use the identity \( \sin^3 x = \frac{3\sin x - \sin 3x}{4} \): \( \int \sin^3 x \cos^2 x \, dx = \int \frac{3\sin x - \sin 3x}{4} \cos^2 x \, dx \). After expansion and simplification (or using substitution \( u = \cos x \), \( du = -\sin x \, dx \)): \( \int \sin^3 x \cos^2 x \, dx = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} \). For the second integral, use integration by parts with \( u = x \), \( dv = e^x dx \): \( \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x = e^x(x - 1) \). Combining: \( y = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + e^x(x - 1) + C \).

Exam Tip: For direct integration of sums, split the integral and handle each term individually; this organization prevents algebraic errors and makes the solution readable.

Question 27. Find the general solution of the differential equation: \( y \, dx + (1 + x^2) \tan^{-1} x \, dy = 0 \)
Answer: Rearrange: \( y \, dx = -(1 + x^2) \tan^{-1} x \, dy \). Separate: \( \frac{dx}{(1 + x^2) \tan^{-1} x} = -\frac{dy}{y} \). Integrate both sides: \( \int \frac{dx}{(1 + x^2) \tan^{-1} x} = -\int \frac{dy}{y} \). For the left integral, let \( u = \tan^{-1} x \), so \( du = \frac{dx}{1 + x^2} \): \( \int \frac{du}{u} = \log|\tan^{-1} x| \). For the right integral: \( -\log|y| \). Combining: \( \log|\tan^{-1} x| = -\log|y| + \log C \), which simplifies to: \( \tan^{-1} x \cdot y = C \).

Exam Tip: When the denominator contains a composition like \( (1 + x^2) f(x) \) where \( f'(x) = \frac{1}{1 + x^2} \), let \( u = f(x) \) to reveal the logarithmic integral \( \int \frac{du}{u} \).

Question 28. Find the particular solution of the differential equation \( \frac{dy}{dx} = 1 + x + y + xy \), given that y = 0 when x = 1.
Answer: Factor: \( \frac{dy}{dx} = (1 + x)(1 + y) \). Separate: \( \frac{dy}{1 + y} = (1 + x) dx \). Integrate: \( \log|1 + y| = x + \frac{x^2}{2} + C \). Substitute \( y = 0, x = 1 \): \( 0 = 1 + \frac{1}{2} + C \), so \( C = -\frac{3}{2} \). The solution is: \( \log|1 + y| = x + \frac{x^2}{2} - \frac{3}{2} \), which gives \( y = e^{x + \frac{x^2}{2} - \frac{3}{2}} - 1 \).

Exam Tip: Always factor the right side of the differential equation immediately; this is often the key step that enables variable separation and reveals the structure of the problem.

Question 41. Find the general solution of the differential equation \( \frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x} \)
Answer: We have \( \frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x} \)

Separating variables: \( dy = \frac{1 - \cos x}{1 + \cos x} dx \)

Using the identity \( \cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \):

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

\( dy = \frac{\left[1 + \tan^2(x/2)\right] - \left[1 - \tan^2(x/2)\right]}{1 + \tan^2(x/2) + 1 - \tan^2(x/2)} \cdot \frac{1 + \tan^2(x/2)}{1 + \tan^2(x/2)} dx \)

\( dy = \frac{2\tan^2(x/2)}{2} dx \)

\( dy = \tan^2(x/2) dx \)

Integrating both sides:

\( \int dy = \int \tan^2(x/2) dx \)

\( y = \int \left[\sec^2(x/2) - 1\right] dx \)

Using the formula \( \sec^2 x - \tan^2 x = 1 \) and \( \frac{d}{dx}\tan(x/2) = \sec^2(x/2) \cdot \frac{1}{2} \):

\( y = 2\tan(x/2) - x + C \)
In simple words: Change the equation into a form where you can separate x and y on opposite sides. Use trigonometric identities to simplify, then integrate each side. The final answer contains tangent of half the angle, minus the variable, plus a constant.

Exam Tip: Recognize half-angle identities early - they simplify complex trigonometric fractions significantly. Always verify your answer by differentiating it back to check it matches the original equation.

Question 42. Solve the differential equation \( (x^2 - yx^2)dy + (y^2 + x^2y^2) dx = 0 \), given that \( y = 1 \) when \( x = 1 \).
Answer: We have \( x^2(1 - y)dy + y^2(1 + x^2)dx = 0 \)

Dividing by \( x^2y^2 \):

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

Integrating both sides:

\( \int \frac{(1-y)}{y^2} dy + \int \frac{(1+x^2)}{x^2} dx = 0 \)

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

For \( y = 1, x = 1 \):

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

\( c = -1 \)

Hence, the required solution is:

\( \frac{1}{y} + \log y + \frac{1}{x} - x = 1 \)
In simple words: Factor out common terms from the equation, then separate the variables. Integrate each part separately. Use the given point to find the constant of integration.

Exam Tip: Always factor first before separating variables - it simplifies the algebra significantly. When applying initial conditions, substitute carefully to avoid arithmetic errors.

Question 43. Find the particular solution of the differential equation \( e^x\sqrt{1 - y^2}dx + \frac{y}{x}dy = 0 \), given that \( y = 1 \) when \( x = 0 \).
Answer: Given: \( e^x\sqrt{1 - y^2}dx + \frac{y}{x}dy = 0 \)

Separating variables:

\( xe^x dx + \frac{y}{\sqrt{1-y^2}} dy = 0 \)

\( \int xe^x dx + \int \frac{y}{\sqrt{1-y^2}} dy = 0 \)

Substituting \( \sqrt{1 - y^2} = t \), so \( 1 - y^2 = t^2, -2ydy = 2tdt \):

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

For \( y = 1 \) and \( x = 0 \):

\( 0 - 1 - 0 = c \)

\( c = -1 \)

Hence, the particular solution is:

\( xe^x - e^x - \frac{1}{2}\log|\sqrt{1 - y^2}| + 1 = 0 \)
In simple words: Rearrange to get each variable on its own side. Integrate using substitution where needed. Apply the starting condition to determine the constant.

Exam Tip: Use substitution only when the integral resists standard formulas. Always track the substitution bounds or back-substitute carefully to avoid sign errors.

Question 44. Find the particular solution of the differential equation \( \frac{dy}{dx} = \frac{x(2\log x + 1)}{(\sin y + y\cos y)} \), given that \( y = \frac{\pi}{2} \) when \( x = 1 \).
Answer: Given: \( \frac{dy}{dx} = \frac{x(2\log x + 1)}{(\sin y + y\cos y)} \)

\( \int \sin y dy + \int y\cos y dy = \int 2x\log x dx + \int x dx \)

Let \( \int y\cos y dy = I \). Then:

\( \int y\cos y dy = \left(\int \cos y dy\right) y - \int \left(\int \cos y dy\right) \cdot \frac{d}{dx}y dy \)

And \( \int x\log x = \left(\int x dx\right)\log x - \int \left(\int x dx\right) \frac{d}{dx}\log x dx \)

We have:

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

For \( y = \frac{\pi}{2}, x = 1 \):

\( 0 + \frac{\pi}{2} + 0 = 0 - 0 + 0 + c \)

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

\( y\sin y + \cos y = x^2\log x + \frac{\pi}{2} \)
In simple words: Rearrange the equation to isolate each variable. Use integration by parts for products of functions. Substitute the initial values to find the constant.

Exam Tip: Integration by parts appears frequently - practice the formula \( \int u \, dv = uv - \int v \, du \) until it becomes automatic. Always check initial conditions carefully, especially when they involve special angles.

Question 45. Solve the differential equation \( \frac{dy}{dx} = y\sin 2x \), given that \( y(0) = 1 \).
Answer: We have \( \frac{dy}{dx} = y\sin 2x \)

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

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

For \( y = 1, x = 0 \):

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

\( \log y = \frac{1}{2}(1 - \cos 2x) \)

\( \log y = \sin^2 x \)

Thus, the particular solution is:

\( y = e^{\sin^2 x} \)
In simple words: Separate y and x to opposite sides. Integrate both sides. Use the starting condition to find the constant. Simplify using trigonometric identities.

Exam Tip: When you obtain \( \log y \), always exponentiate carefully to get y on its own. Watch for opportunities to use the identity \( 1 - \cos 2x = 2\sin^2 x \).

Question 46. Solve the differential equation \( (x + 1)\frac{dy}{dx} = 2xy \), given that \( y(2) = 3 \).
Answer: Given: \( (x + 1)\frac{dy}{dx} = 2xy \)

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

\( \log y = \int 2 - \frac{2}{x+1} dx \)

\( \log y = 2x - 2\log(x + 1) + c \)

For \( x = 2 \) and \( y = 3 \):

\( c = 3\log 3 - 4 \)

Hence, the particular solution is:

\( y(x + 1)^2 = 27 \)
In simple words: Separate the variables onto opposite sides. Integrate using polynomial division to handle the fraction. Use the given point to find the constant.

Exam Tip: When the integral contains a rational function, use polynomial long division to split it into simpler parts. This often produces a logarithm plus a polynomial term.

Question 47. Solve \( \frac{dy}{dx} = x(2\log x + 1) \), given that \( y = 0 \) when \( x = 2 \).
Answer: We have \( \frac{dy}{dx} = 2x\log x + x \). Integrating:

\( y = \int (2x\log x + x)dx \)

\( y = \int 2x\log x dx + \int x dx \)

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

Given that \( y = 0 \) when \( x = 2 \):

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

Now putting \( x = 2 \) and \( y = 0 \):

\( 0 = 4\log 2 + c \)

\( c = -4\log 2 \)

Thus, the solution is:

\( y = x^2\log x - 4\log 2 \)
In simple words: Integrate term by term. Use integration by parts for the product of x and logarithm. Substitute the starting values to find the constant.

Exam Tip: Integration by parts on logarithmic terms requires careful setup - keep the logarithm as u and the polynomial as dv. Always expand the full antiderivative before applying initial conditions.

Question 48. Solve \( (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have \( (x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x \)

Given that: \( y = 1 \) when \( x = 0 \)

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

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

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

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

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

\( y = \frac{3}{4}\log|x^2 + 1| + \log|x + 1| - \frac{1}{2}\tan^{-1} x + \frac{1}{2}\log|x + 1| + c \)

\( y = \frac{3}{4}\log|x^2 + 1| + \frac{1}{2}\log|x + 1| - \frac{1}{2}\tan^{-1} x + c \)

For \( y = 1, x = 0 \):

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

\( c = 1 \)
In simple words: Factor the denominator completely. Use partial fractions to break the rational function into simpler pieces. Integrate each piece separately, then use the initial condition.

Exam Tip: Partial fraction decomposition is key here - factor the denominator first, then match coefficients carefully. Always verify your decomposition by recombining the fractions.

Question 49. Solve \( \frac{dy}{dx} = y\tan x \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have \( \frac{dy}{dx} = y\tan x \)

Given that: \( y = 1 \) when \( x = 0 \)

\( \frac{dy}{y} = y\tan x \)

\( \frac{dy}{y} = \tan x dx \)

\( \log y = \log|\sec x| + c \)

\( 0 = 0 + c \)

\( y\cos x = 1 \) is the particular solution
In simple words: Separate variables by moving all y terms to one side and all x terms to the other. Integrate each side. Apply the starting condition to eliminate the constant.

Exam Tip: Recognize that \( \int \tan x \, dx = \log|\sec x| + C \) is a standard integral. When you get \( \log y = \log|\sec x| \), exponentiate immediately to get \( y = |\sec x| \), then simplify using the initial condition.

Question 50. Solve \( \frac{dy}{dx} = y^2 \tan 2x \), given that \( y = 2 \) when \( x = 0 \).
Answer: We have: \( \frac{dy}{dx} = y^2 \tan 2x \)

Given that, \( y = 2 \) when \( x = 0 \)

\( \frac{dy}{y^2} = \tan 2x dx \)

\( \int \frac{dy}{y^2} = \int \tan 2x dx \) (integrating both sides)

\( -\frac{1}{y} = \frac{\log(\sec 2x)}{2} \)

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

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

\( y(1 + \log\cos 2x) = 2 \) (is the particular solution)
In simple words: Separate y and x onto opposite sides. Integrate each side separately. Apply the starting values to determine the constant of integration.

Exam Tip: When integrating \( \tan 2x \), remember to account for the chain rule factor - the integral is \( \frac{\log|\sec 2x|}{2} \), not \( \log|\sec 2x| \). Check your answer by differentiating.

Question 51. Solve \( \frac{dy}{dx} = y\cot 2x \), given that \( y = 2 \) when \( x = \frac{\pi}{4} \).
Answer: We have \( \frac{dy}{dx} = y\cot 2x \)

Given that, \( y = 2 \) when \( x = \frac{\pi}{2} \)

\( \frac{dy}{y} = y\cot 2x \)

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

\( \int \frac{dy}{y} = \int \cot 2x dx \)

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

\( \log 2 = 0 + c \)

\( c = \log 2 \)

Thus, the particular solution is:

\( \sec y = \sqrt{2\cos x} \)
In simple words: Separate the variables. Integrate using the standard formula for cotangent. Use the given point to find the constant.

Exam Tip: The integral of \( \cot x \) is \( \log|\sin x| \) - remember that cotangent has sine in the numerator when expressed in logarithmic form. Always verify special angle values carefully.

Question 52. Solve \( (1 + x^2)\sec^2 y dy + 2x\tan y dx = 0 \), given that \( y = \frac{\pi}{4} \) when \( x = 1 \).
Answer: We have, \( (1 + x^2)\sec^2 y dy + 2x\tan y dx = 0 \)

Given that, when \( x = 1 \), \( y = \frac{\pi}{4} \)

\( (1 + x^2)\sec^2 y dy + 2x\tan y dx = 0 \)

\( \frac{\sec^2 y}{\tan y} dy + \frac{2x}{1 + x^2} dx = 0 \)

\( \int \frac{\sec^2 y}{\tan y} dy + \int \frac{2x}{1 + x^2} dx = 0 \)

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

For \( y = \frac{\pi}{4}, x = 1 \)

We have, \( 0 + \log 2 = \log c \)

\( c = 2 \)

Hence the required particular solution is:

\( \tan y(1 + x^2) = 2 \)
In simple words: Rearrange the equation to isolate terms involving y and x. Integrate both sides. Apply the given condition to calculate the constant.

Exam Tip: When you obtain a product of logarithms, combine them using log properties before exponentiating. This keeps the final answer in its simplest implicit form.

Question 53. Find the equation of the curve passing through the point \( \left(0, \frac{\pi}{4}\right) \) whose differential equation is \( \sin x\cos y dx + \cos x\sin y dy = 0 \).
Answer: We have, \( \sin x\cos y dx + \cos x\sin y dy = 0 \)

\( \sin x\cos y dx + \cos x\sin y dy = 0 \)

\( \tan x dx + \tan y dy = 0 \)

\( \log|\sec x| + \log|\sec y| = \log c \)

\( \sec x\sec y = c \)

Given that, coordinates of point \( \left(0, \frac{\pi}{4}\right) \)

\( c = \sqrt{2} \)

\( \sec y = \sqrt{2}\cos x \)

\( y = \cos^{-1}\left(\frac{1}{\sqrt{2}\cos x}\right) \) (is the required particular solution)
In simple words: Divide through to separate variables into x and y terms. Integrate each side. Substitute the given point to find the constant of integration.

Exam Tip: Products of sines and cosines often factor into products of tangents after dividing by appropriate trigonometric terms. Always verify that your solution passes through the given point before finalizing.

Question 54. Find the equation of a curve which passes through the origin and whose differential equation is \( \frac{dy}{dx} = e^x\sin x \).
Answer: Given, \( \frac{dy}{dx} = e^x\sin x \)

\( dy = e^x\sin x dx \)

\( \int dy = \int e^x\sin x dx \)

Let \( I = \int e^x\sin x dx \)

\( I = \int e^x dx\sin x - \int \left(\int e^x dx\right) \frac{d}{dx}\sin x dx \)

\( I = e^x\sin x - \int e^x\cos x dx \)

\( I = e^x\sin x - \int e^x dx\cos x - \int \left(\int e^x dx\right) \frac{d}{dx}\cos x dx \)

\( I = e^x\sin x - e^x\cos x - \int e^x\sin x dx \)

\( 2I = e^x\sin x - e^x\cos x \)

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

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

For the curve passes through (0,0)

We have, \( c = \frac{1}{2} \)

\( 2y = e^x\sin x + e^x\cos x = 1 \)
In simple words: Integrate directly. For the product of exponential and trigonometric functions, use integration by parts twice, then solve for the integral. Apply the origin to find the constant.

Exam Tip: The double-application integration by parts technique creates an equation in I itself - solve for I algebraically rather than continuing to expand. This saves time and reduces error.

Question 55. A curve passes through the point (0, -2) and at any point (x, y) of the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. Find the equation of the curve.
Answer: Given that the product of slope of tangent and y coordinate equals the x-coordinate, i.e., \( y\frac{dy}{dx} = x \)

We have, \( ydy = xdx \)

\( \int ydy = \int xdx \)

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

For the curve passes through (0, -2), we get \( c = 2 \)

Thus, the required particular solution is:

\( y^2 = x^2 + 4 \)
In simple words: Translate the word problem into a differential equation. Separate variables. Integrate both sides. Apply the given point to determine the constant.

Exam Tip: Word problems require careful translation - identify what "the product of... and..." means mathematically. Always keep track of which variable is which during translation.

Question 56. A curve passes through the point (-1, 1) and at any point (x, y) of the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve.
Answer: Given: \( \frac{dy}{dx} = 2 \cdot \frac{y - (-3)}{x - (-4)} \)

\( \frac{dy}{dx} = \frac{2(y + 3)}{x + 4} \)

\( \frac{dy}{y + 3} = \frac{2dx}{x + 4} \)

\( \int \frac{dy}{y + 3} = 2\int \frac{dx}{x + 4} \)

\( \log(y + 3) = 2\log(x + 4) + c \)

The curve passes through (-2, 1), we have, \( c = 0 \)

\( y + 3 = (x + 4)^2 \)
In simple words: Express the geometric condition as a differential equation. Separate and integrate. Use the given point to find the constant.

Exam Tip: When comparing slopes, remember that the slope of a line segment between two points is \( \frac{\Delta y}{\Delta x} \). Always simplify logarithmic equations by exponentiating both sides before finalizing.

Question 57. In a bank, principal increases at the rate \( r\% \) per annum. Find the value of r if Rs 100 doubles itself in 10 years. (Given \( \log_e 2 = 0.6931 \))
Answer: Given: \( \frac{dp}{dt} = \left(\frac{r}{100}\right) \times p \)

Here, p is the principal, r is the rate of interest per annum and t is the time in years.

Solving the differential equation we get:

\( \frac{dp}{p} = \left(\frac{r}{100}\right) dt \)

\( \int \frac{dp}{p} = \int \frac{r}{100} dt \)

\( \log p = \frac{rt}{100} + c \)

\( p = e^{\frac{rt}{100} + c} \)

As it is given that the principal doubles itself in 10 years, so

Let the initial interest be p1 (for t = 0), after 10 years p1 becomes 2p1.

Thus, for (t = 0): \( p1 = e^c \) ...(i)

\( p = 2p1 = e^{\frac{r(10)}{100}} \cdot e^c \) ...(ii)

Substituting (i) in (ii), we get:

\( 2p1 = e^{\frac{r}{10}} \cdot p1 \)

\( 2 = e^{\frac{r}{10}} \)

\( \log 2 = \frac{r}{10} \)

\( r = 10\log 2 \)

\( r = 6.931 \)

Rate of interest = 6.931
In simple words: Set up the differential equation for exponential growth. Solve by separating variables and integrating. Use the doubling condition to find the rate.

Exam Tip: Exponential growth/decay problems always have the form \( \frac{dA}{dt} = kA \). The solution is always \( A = A_0 e^{kt} \). Use the given conditions to find k quickly without writing intermediate steps.

Question 58. In a bank, principal increases at the rate of 5% per annum. An amount of Rs 1000 is deposited in the bank. How much will it worth after 10 years? (Given \( e^{0.5} = 1.648 \))
Answer: Given: rate of interest = 5%

P(initial) = Rs 1000

And, \( \frac{dp}{dt} = \frac{5}{100} \times p \)

\( \frac{dp}{p} = \frac{5}{100} dt \)

\( \int \frac{dp}{p} = \int \frac{5}{100} dt \)

\( \log p = \frac{5t}{100} + c \)

\( p = e^{\frac{5t}{100} + c} \)

For t = 0, we have p = 1000

\( 1000 = e^c \)

For t = 10 years we have, \( p = e^{\frac{50}{100}}.1000 \)

\( p = 1000e^{1/2} \)

\( p = 1648 \)

Thus, principal is Rs 1648 for t = 10 years.
In simple words: Write the exponential growth equation. Solve by integration. Apply the initial condition. Substitute the time value and the given exponential value to find the final amount.

Exam Tip: When working with percentage rates, convert to decimal form immediately - 5% becomes 0.05 or 5/100. Always identify the initial condition (at t=0) to fix the integration constant before answering the main question.

Question 59. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Answer: Given:

Volume \( V = \frac{4\pi r^3}{3} \)

\( \frac{dV}{dt} = \frac{4}{3}\pi 3r^2 \frac{dr}{dt} \)

\( \frac{dr}{dt} = k \) (constant)

\( 4\pi r^2 \frac{dr}{dt} = k \)

\( 4\pi r^2 dr = k dt \)

\( \int 4\pi r^2 dr = \int k dt \)

\( \frac{4\pi r^3}{3} = kt + c \)

For t = 0, r = 3 and for t = 3, r = 6. So, we have:

\( \frac{4\pi (3)^3}{3} = 0 + c \)

\( c = 36\pi \)

\( \frac{4\pi (6)^3}{3} = k(3) + 36\pi \)

\( k = 84\pi \)

So after t seconds the radius of the balloon will be:

\( \frac{4\pi r^3}{3} = 84\pi t + 36\pi \)

\( 4\pi r^3 = 252\pi t + 108\pi \)

\( r^3 = \frac{252\pi t + 108\pi}{4\pi} \)

\( r^3 = 63t + 27 \)

\( r = \sqrt[3]{63t + 27} \)

Hence, radius of the balloon as a function of time is \( r = (63t + 27)^{1/3} \)
In simple words: Differentiate the volume formula to relate the rate of volume change to the rate of radius change. Integrate. Use the two given conditions to find both the constant of integration and the rate constant.

Exam Tip: Related rates problems require careful differentiation of the geometric formula. Always identify what varies with time (usually a dimension or volume) versus what stays constant. Use both given conditions systematically - one fixes the constant, the other fixes the rate.

Question 60. In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?
Answer: Let y be the bacteria count, then, we have:

Rate of growth of bacteria is proportional to the number present

So, \( \frac{dy}{dt} = cy \)

Where c is a constant,

Then, solving the equation we have:

\( \frac{dy}{y} = cdt \)

\( \log y = ct + k \)

Where k is constant of integration

\( y = e^{ct + k} \)

And we have for t = 0, y = 10000:

\( 10000 = e^k \) ...(i)

For t = 2 hrs, y is increased by 10%, i.e., y = 110000

\( 110000 = e^{c(2)} \cdot e^k \)

\( 110000 = e^{2c}(10000) \) (from (i))

\( e^{2c} = 1.1 \)

\( c = \frac{1}{2}\log\left(\frac{11}{10}\right) \)

When y = 200000, we have:

\( 200000 = e^{ct} \cdot 100000 \)

\( e^{ct} = 2 \)

\( (e^c)^t = 2 \)

\( tc = \log 2 \)

\( t = \frac{2\log 2}{\log\frac{11}{10}} \)

Hence, \( t = \frac{2\log 2}{\log\frac{11}{10}} \)
In simple words: Write the proportionality condition as a differential equation. Solve by separation of variables. Use the first condition to find the integration constant, then use the second condition to find the growth rate. Finally, apply the target population to find the required time.

Exam Tip: Proportional growth always yields an exponential model. The percentage increase after a fixed time (10% in 2 hours) is the key to finding the growth constant - use it to calculate \( e^{kT} \) directly. Keep logarithmic expressions intact until the final simplification unless a numerical answer is explicitly requested.