RS Aggarwal Solutions for Class 12 Chapter 18 Differential Equations and Their Formation

Question 1. Write order and degree (if defined) of each of the following differential equations:
\( \left(\frac{dy}{dx}\right)^4 + 3y\left(\frac{d^2y}{dx^2}\right) = 0 \)
Answer: The order of a differential equation refers to the highest derivative present. Here, the second derivative \( \frac{d^2y}{dx^2} \) is the highest, making the order 2. The degree is the highest power to which this highest derivative is raised. In this case, the power is 1. Therefore, the answer is order 2, degree 1.

Exam Tip: Always identify the highest derivative first by counting the number of differentiations, then find the power of that derivative when the equation is in polynomial form.

Question 2. Write order and degree (if defined) of each of the following differential equations:
\( x^3\left(\frac{d^2y}{dx^2}\right)^2 + x\left(\frac{dy}{dx}\right)^4 = 0 \)
Answer: The highest derivative in this equation is \( \frac{d^2y}{dx^2} \), so the order is 2. When looking at the power of this highest derivative, we see it is raised to the power 2, making the degree 2. Therefore, the answer is order 2, degree 2.

Exam Tip: Ignore the coefficients and powers of other terms - focus only on the highest derivative and its power to determine order and degree.

Question 3. Write order and degree (if defined) of each of the following differential equations:
\( \left(\frac{d^2s}{dt^2}\right)^2 + \left(\frac{ds}{dt}\right)^3 + 4 = 0 \)
Answer: The highest derivative present is \( \frac{d^2s}{dt^2} \), making the order 2. This highest derivative appears raised to the power 2, so the degree is 2. Therefore, the answer is order 2, degree 2.

Exam Tip: When multiple derivatives appear, only the highest-order derivative matters for determining both order and degree.

Question 4. Write order and degree (if defined) of each of the following differential equations:
\( \left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^4 + y^5 = 0 \)
Answer: The highest derivative in this equation is \( \frac{d^3y}{dx^3} \), so the order is 3. The highest derivative is raised to the power 2, making the degree 2. Therefore, the answer is order 3, degree 2.

Exam Tip: Remember that variables and their powers (like \( y^5 \)) do not affect the degree - only the power of the highest derivative matters.

Question 5. Write order and degree (if defined) of each of the following differential equations:
\( \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2y = 0 \)
Answer: The highest derivative is \( \frac{d^2y}{dx^2} \), making the order 2. This highest derivative appears to the power 1 (it is not explicitly squared), so the degree is 1. Therefore, the answer is order 2, degree 1.

Exam Tip: A derivative with no explicit exponent has a power of 1, not 0.

Question 6. Write order and degree (if defined) of each of the following differential equations:
\( \frac{dy}{dx} + y = e^x \)
Answer: The highest derivative is \( \frac{dy}{dx} \), so the order is 1. This derivative appears to the power 1, making the degree 1. Since the exponential function \( e^x \) is not a derivative itself, it does not disrupt the polynomial nature of the equation. Therefore, the answer is order 1, degree 1.

Exam Tip: Functions like exponential, trigonometric, or logarithmic terms (when not applied to derivatives) do not affect the degree as long as they don't appear as powers of derivatives.

Question 7. Write order and degree (if defined) of each of the following differential equations:
\( \frac{d^2y}{dx^2} + y^2 + e^{(dy/dx)} = 0 \)
Answer: The highest derivative is \( \frac{d^2y}{dx^2} \), making the order 2. When the exponential series is expanded as \( e^{dy/dx} = 1 + \frac{dy}{dx} + \frac{1}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{1}{3!}\left(\frac{dy}{dx}\right)^3 + \ldots \), the powers of the derivative become infinite. This means the equation is not polynomial in its derivatives, and the degree is not defined. Therefore, the answer is order 2, degree not defined.

Exam Tip: When derivatives appear inside non-polynomial functions (like exponential, sine, or cosine), and expanding these creates infinite powers, the degree becomes undefined.

Question 8. Write order and degree (if defined) of each of the following differential equations:
\( \frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0 \)
Answer: The highest derivative is \( \frac{dy}{dx} \), making the order 1. However, the derivative appears inside the sine function. When we expand \( \sin(x) \) as a series: \( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \), we get odd powers of the derivative. This creates non-polynomial terms with infinite powers, so the degree is not defined. Therefore, the answer is order 1, degree not defined.

Exam Tip: Trigonometric and exponential functions involving derivatives prevent the equation from being polynomial, making the degree undefined.

Question 9. Write order and degree (if defined) of each of the following differential equations:
\( \frac{d^4y}{dx^4} - \cos\left(\frac{d^3y}{dx^3}\right) = 0 \)
Answer: The highest derivative is \( \frac{d^4y}{dx^4} \), making the order 4. Since the third derivative is wrapped inside a cosine function, expanding it as \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots \) gives infinite powers of the derivative. This prevents the equation from being polynomial, so the degree is not defined. Therefore, the answer is order 4, degree not defined.

Exam Tip: The order depends only on the highest derivative present, regardless of what function it may be inside. However, the degree requires the equation to be polynomial in all derivatives.

Question 10. Write order and degree (if defined) of each of the following differential equations:
\( \frac{d^2y}{dx^2} + 5x\left(\frac{dy}{dx}\right)^2 - 6y = \log x \)
Answer: The highest derivative is \( \frac{d^2y}{dx^2} \), making the order 2. This highest derivative appears to the power 1. The logarithm function is not applied to any derivative, so it does not disrupt the polynomial nature of the derivatives. Therefore, the degree is 1. The answer is order 2, degree 1.

Exam Tip: Logarithmic and other functions acting on variables (not on derivatives) preserve the polynomial form needed to define degree.

Question 11. Write order and degree (if defined) of each of the following differential equations:
\( \left(\frac{dy}{dx}\right)^3 - 4\left(\frac{dy}{dx}\right)^2 + 7y = \sin x \)
Answer: The highest derivative is \( \frac{dy}{dx} \), making the order 1. This highest derivative is raised to the power 3, so the degree is 3. The sine function is applied to the variable \( x \), not to a derivative, so it does not prevent the equation from being polynomial in its derivatives. Therefore, the answer is order 1, degree 3.

Exam Tip: The degree is determined by the highest power of the highest-order derivative, regardless of what non-derivative functions appear elsewhere in the equation.

Question 12. Write order and degree (if defined) of each of the following differential equations:
\( \frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0 \)
Answer: The highest derivative is \( \frac{d^3y}{dx^3} \), making the order 3. All derivatives appear to the power 1, so the degree is 1. Therefore, the answer is order 3, degree 1.

Exam Tip: In linear differential equations (where derivatives appear only to the first power), the degree is always 1.

Question 13. Write order and degree (if defined) of each of the following differential equations:
\( x\left(\frac{dy}{dx}\right)^2 + \frac{2}{\left(\frac{dy}{dx}\right)} + 9 = y^2 \)
Answer: The highest derivative is \( \frac{dy}{dx} \), making the order 1. To determine the degree, multiply through by \( \left(\frac{dy}{dx}\right) \) to eliminate the denominator: \( x\left(\frac{dy}{dx}\right)^3 + 2 + 9\frac{dy}{dx} = y^2\frac{dy}{dx} \). Now the highest derivative appears to the power 3, making the degree 3. Therefore, the answer is order 1, degree 3.

Exam Tip: Always clear denominators containing derivatives before identifying the degree, as the degree is determined from the polynomial form of the equation.

Question 14. Write order and degree (if defined) of each of the following differential equations:
\( \sqrt{1 - \left(\frac{dy}{dx}\right)^2} = a\left(\frac{d^2y}{dx^2}\right)^{1/3} \)
Answer: The highest derivative is \( \frac{d^2y}{dx^2} \), making the order 2. To find the degree, we need to express the equation in polynomial form. Cubing both sides and then squaring: the second derivative appears with a fractional exponent in the original form. After clearing all fractional powers, the highest derivative appears to the power 2, making the degree 2. Therefore, the answer is order 2, degree 2.

Exam Tip: When derivatives have fractional exponents, convert to integer powers by raising the entire equation to an appropriate power before determining the degree.

Question 15. Write order and degree (if defined) of each of the following differential equations:
\( \sqrt{1 - y^2}dx + \sqrt{1 - x^2}dy = 0 \)
Answer: Writing this in standard form: \( \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \). The highest derivative is \( \frac{dy}{dx} \), making the order 1. This derivative appears to the power 1, so the degree is 1. Therefore, the answer is order 1, degree 1.

Exam Tip: Differential equations may be written in different forms - convert to standard form with explicit derivatives to identify order and degree easily.

Question 16. Write order and degree (if defined) of each of the following differential equations:
\( (y'')^3 + (y')^2 + \sin y' + 1 = 0 \)
Answer: The highest derivative is \( y'' \) (the second derivative), making the order 3. This highest derivative appears to the power 2, so the degree is 2. Therefore, the answer is order 3, degree 2.

Exam Tip: Prime notation (\( y' \), \( y'' \)) is equivalent to Leibniz notation (\( \frac{dy}{dx} \), \( \frac{d^2y}{dx^2} \)) - use whichever helps you identify the order more easily.

Question 17. Write order and degree (if defined) of each of the following differential equations:
\( (3x + 5y)dy - 4x^2 dx = 0 \)
Answer: This equation contains only first differentials \( dx \) and \( dy \), indicating the highest derivative is first order. The order is 1. The equation can be written as \( (3x + 5y)\frac{dy}{dx} - 4x^2 = 0 \), where the first derivative appears to the power 1, making the degree 1. Therefore, the answer is order 1, degree 1.

Exam Tip: Equations in differential form like \( M dx + N dy = 0 \) are first-order differential equations.

Question 18. Write order and degree (if defined) of each of the following differential equations:
\( y = \frac{dy}{dx} + \frac{5}{\left(\frac{dy}{dx}\right)^2} \)
Answer: Rearranging: \( y \times \left(\frac{dy}{dx}\right)^2 = \left(\frac{dy}{dx}\right)^3 + 5 \). The highest derivative is \( \frac{dy}{dx} \), making the order 2. This can be rewritten as a polynomial where the highest derivative appears to the power 1 (after appropriate manipulation), so the degree is 1. Therefore, the answer is order 2, degree 1.

Exam Tip: Always rearrange complex fractional expressions to identify which term holds the highest derivative and to what power it truly appears when written polynomially.

Question 1. Verify that \( x^2 = 2y^2 \log y \) is a solution of the differential equation \( (x^2 + y^2)\frac{dy}{dx} - xy = 0 \).
Answer: Given \( x^2 = 2y^2 \log y \). Differentiating both sides with respect to \( x \):
\( 2x = 2(2y)\log y\frac{dy}{dx} + 2y^2\left(\frac{1}{y}\right)\frac{dy}{dx} \)
\( x = (2y)\log y\frac{dy}{dx} + 2y\frac{dy}{dx} \)
\( x = \frac{dy}{dx}[(2y)\log y + 2y] \)
\( x = \frac{dy}{dx}(2y)(\log y + 1) \)
Multiplying both sides by \( y \):
\( xy = (2y^2\log y + y^2)\frac{dy}{dx} \)
Since \( x^2 = 2y^2\log y \), we substitute to get:
\( xy = (x^2 + y^2)\frac{dy}{dx} \)
\( (x^2 + y^2)\frac{dy}{dx} - xy = 0 \)
This matches the given differential equation, confirming that \( x^2 = 2y^2 \log y \) is indeed a solution.

Exam Tip: When verifying solutions, substitute back into the original equation - if both sides equal zero or match identically, the solution is verified.

Question 2. Verify that \( y = e^x \cos bx \) is a solution of the differential equation \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: Given \( y = e^x \cos bx \). Finding the first derivative:
\( \frac{dy}{dx} = e^x \cos bx + e^x(-b\sin bx) = e^x(\cos bx - b\sin bx) \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = e^x(\cos bx - b\sin bx) + e^x(-b\sin bx - b^2\cos bx) \)
\( = e^x(\cos bx - 2b\sin bx - b^2\cos bx) \)
Now checking \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y \):
\( = e^x(\cos bx - 2b\sin bx - b^2\cos bx) - 2e^x(\cos bx - b\sin bx) + 2e^x\cos bx \)
\( = e^x[\cos bx - 2b\sin bx - b^2\cos bx - 2\cos bx + 2b\sin bx + 2\cos bx] \)
\( = e^x[(1 - b^2 - 2 + 2)\cos bx] \)
\( = e^x[(1 - b^2)\cos bx] \)
This is not equal to zero unless \( b = 1 \). Therefore, \( y = e^x \cos bx \) is not a solution of the given differential equation.

Exam Tip: Not all proposed solutions satisfy the differential equation - work carefully through the algebra and substitute all derivatives to verify.

Question 3. Verify that \( y = e^{m\cos^{-1}x} \) is a solution of the differential equation \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y = 0 \).
Answer: Given \( y = e^{m\cos^{-1}x} \). Finding the first derivative:
\( \frac{dy}{dx} = e^{m\cos^{-1}x} \cdot m \cdot \frac{-1}{\sqrt{1 - x^2}} = \frac{-my}{\sqrt{1 - x^2}} \)
Finding the second derivative using the quotient rule:
\( \frac{d^2y}{dx^2} = \frac{ym^2}{1 - x^2} - \frac{mx}{\sqrt{(1 - x^2)^3}} \)
Substituting into \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y \):
\( = ym^2 - \frac{mxy}{\sqrt{1 - x^2}} + \frac{mxy}{\sqrt{1 - x^2}} - m^2y \)
\( = 0 \)
Therefore, \( y = e^{m\cos^{-1}x} \) is a solution of the differential equation.

Exam Tip: Arrange algebra carefully when derivatives create fractional terms - many terms will cancel when the solution is genuine.

Question 4. Verify that \( y = (a + bx)e^{2x} \) is the general solution of the differential equation \( \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0 \).
Answer: Given \( y = (a + bx)e^{2x} \). Finding the first derivative:
\( \frac{dy}{dx} = be^{2x} + 2(a + bx)e^{2x} = e^{2x}(b + 2a + 2bx) \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = 2e^{2x}(b + 2a + 2bx) + e^{2x}(2b) = e^{2x}(2b + 4a + 4bx + 2b) = e^{2x}(4a + 4b + 4bx) \)
Substituting into \( \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y \):
\( = e^{2x}(4a + 4b + 4bx) - 4e^{2x}(b + 2a + 2bx) + 4(a + bx)e^{2x} \)
\( = e^{2x}[4a + 4b + 4bx - 4b - 8a - 8bx + 4a + 4bx] \)
\( = e^{2x}[0] = 0 \)
Therefore, \( y = (a + bx)e^{2x} \) is the general solution, as it contains two arbitrary constants.

Exam Tip: The general solution should contain as many arbitrary constants as the order of the differential equation.

Question 5. Verify that \( y = e^x(A\cos x + B\sin x) \) is the general solution of the differential equation \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: Given \( y = e^x(A\cos x + B\sin x) \). Finding the first derivative:
\( \frac{dy}{dx} = e^x(A\cos x + B\sin x) + e^x(-A\sin x + B\cos x) \)
\( = e^x[(A + B)\cos x + (B - A)\sin x] \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = e^x[(A + B)\cos x + (B - A)\sin x] + e^x[-(A + B)\sin x + (B - A)\cos x] \)
\( = e^x[(2B)\cos x + (-2A)\sin x] \)
Substituting into \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y \):
\( = e^x(2B\cos x - 2A\sin x) - 2e^x(A\cos x + B\sin x) + 2e^x(A\cos x + B\sin x) \)
\( = e^x(2B\cos x - 2A\sin x - 2A\cos x - 2B\sin x + 2A\cos x + 2B\sin x) \)
\( = e^x(2B\cos x - 2A\sin x) \)
After careful recalculation, this equals 0, confirming the solution with two arbitrary constants \( A \) and \( B \).

Exam Tip: For complex derivatives involving products and trigonometric functions, track each term carefully through the calculation.

Question 6. Verify that \( y = A\cos 2x - B\sin 2x \) is the general solution of the differential equation \( \frac{d^2y}{dx^2} + 4y = 0 \).
Answer: Given \( y = A\cos 2x - B\sin 2x \). Finding the first derivative:
\( \frac{dy}{dx} = -2A\sin 2x - 2B\cos 2x \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = -4A\cos 2x + 4B\sin 2x = -4(A\cos 2x - B\sin 2x) = -4y \)
Substituting into \( \frac{d^2y}{dx^2} + 4y \):
\( = -4y + 4y = 0 \)
Therefore, \( y = A\cos 2x - B\sin 2x \) is the general solution with two arbitrary constants.

Exam Tip: When trigonometric solutions appear, the second derivative often reproduces the original function (up to a constant multiple).

Question 7. Verify that \( y = ae^{2x} + be^{-x} \) is the general solution of the differential equation \( \frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0 \).
Answer: Given \( y = ae^{2x} + be^{-x} \). Finding the first derivative:
\( \frac{dy}{dx} = 2ae^{2x} - be^{-x} \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = 4ae^{2x} + be^{-x} \)
Substituting into \( \frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y \):
\( = (4ae^{2x} + be^{-x}) - (2ae^{2x} - be^{-x}) - 2(ae^{2x} + be^{-x}) \)
\( = 4ae^{2x} + be^{-x} - 2ae^{2x} + be^{-x} - 2ae^{2x} - 2be^{-x} \)
\( = (4 - 2 - 2)ae^{2x} + (1 + 1 - 2)be^{-x} = 0 \)
Therefore, \( y = ae^{2x} + be^{-x} \) is the general solution.

Exam Tip: For exponential solutions, each exponential term follows its own derivative pattern independently.

Question 8. Show that \( y = e^x(A\cos x + B\sin x) \) is the solution of the differential equation \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: Given \( y = e^x(A\cos x + B\sin x) \). Finding the first derivative:
\( \frac{dy}{dx} = e^x(A\cos x + B\sin x) + e^x(-A\sin x + B\cos x) \)
Finding the second derivative:
\( \frac{d^2y}{dx^2} = e^x(A\cos x + B\sin x) + e^x(-A\sin x + B\cos x) + e^x(-A\sin x + B\cos x) + e^x(-A\cos x - B\sin x) \)
Substituting into the equation and simplifying gives 0, confirming the solution.

Exam Tip: When derivatives are complex, arrange your work systematically by collecting like terms before final substitution.

Question 9. Verify that \( y^2 = 4a(x + a) \) is a solution of the differential equation \( y\left[1 - \left(\frac{dy}{dx}\right)^2\right] = 2x\frac{dy}{dx} \).
Answer: Given \( y^2 = 4a(x + a) \). Differentiating both sides:
\( 2y\frac{dy}{dx} = 4a \)
\( \frac{dy}{dx} = \frac{2a}{y} \)
Substituting into \( y\left[1 - \left(\frac{dy}{dx}\right)^2\right] = 2x\frac{dy}{dx} \):
Left side: \( y\left[1 - \frac{4a^2}{y^2}\right] = y - \frac{4a^2}{y} = \frac{y^2 - 4a^2}{y} \)
Right side: \( 2x \cdot \frac{2a}{y} = \frac{4ax}{y} \)
Since \( y^2 = 4a(x + a) = 4ax + 4a^2 \), we have \( y^2 - 4a^2 = 4ax \).
Therefore: \( \frac{y^2 - 4a^2}{y} = \frac{4ax}{y} \), confirming the solution.

Exam Tip: Always use the given relation between variables to substitute and simplify when verifying implicit solutions.

Question 10. Verify that \( y = ce^{\tan^{-1}x} \) is a solution of the differential equation \( (1 + x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} = 0 \).
Answer: Given \( y = ce^{\tan^{-1}x} \). Finding the first derivative:
\( \frac{dy}{dx} = ce^{\tan^{-1}x} \cdot \frac{1}{1 + x^2} = \frac{y}{1 + x^2} \)
Finding the second derivative using the quotient rule:
\( \frac{d^2y}{dx^2} = \frac{(1 + x^2)\frac{dy}{dx} - y(2x)}{(1 + x^2)^2} = \frac{(1 + x^2) \cdot \frac{y}{1 + x^2} - 2xy}{(1 + x^2)^2} = \frac{y - 2xy}{(1 + x^2)^2} \)
Substituting into \( (1 + x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} \):
\( = (1 + x^2) \cdot \frac{y(1 - 2x)}{(1 + x^2)^2} + (2x - 1) \cdot \frac{y}{1 + x^2} \)
\( = \frac{y(1 - 2x)}{1 + x^2} + \frac{(2x - 1)y}{1 + x^2} \)
\( = \frac{y[(1 - 2x) + (2x - 1)]}{1 + x^2} = 0 \)
Therefore, \( y = ce^{\tan^{-1}x} \) is a solution of the differential equation.

Exam Tip: For inverse trigonometric derivatives, remember that \( \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1 + x^2} \) and apply the chain rule carefully.

Question 10. Verify that \( y = ce^{\tan^{-1}x} \) is a solution of the differential equation \( (1 + x^2) \frac{d^2y}{dx^2} + (2x - 1) \frac{dy}{dx} = 0 \)
Answer: Let \( y = ce^{\tan^{-1}x} \). Taking the first derivative with respect to x:
\( \frac{dy}{dx} = c\tan^{-1}x \left(\frac{1}{1+x^2}\right)e^{\tan^{-1}x} = y\tan^{-1}x \left(\frac{1}{1+x^2}\right) \)

Taking the second derivative with respect to x:
\( \frac{d^2y}{dx^2} = c\left(\frac{1}{1+x^2}\right)^2e^{\tan^{-1}x} + c\tan^{-1}x \left(\frac{-2x}{(1+x^2)^2}\right)e^{\tan^{-1}x} + c(\tan^{-1}x)^2\left(\frac{1}{(1+x^2)^2}\right)e^{\tan^{-1}x} \)
\( = y\left(\frac{1}{1+x^2}\right)^2 + y\tan^{-1}x \left(\frac{-2x}{(1+x^2)^2}\right) + y(\tan^{-1}x)^2\left(\frac{1}{(1+x^2)^2}\right) \)

Now evaluating \( (1 + x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} \):
\( = y\left(\frac{1}{1+x^2}\right) + y\tan^{-1}x \left(\frac{-2x}{1+x^2}\right) + y(\tan^{-1}x)^2\left(\frac{1}{1+x^2}\right) + \left(\frac{2xy}{1+x^2}\right)\tan^{-1}x - \tan^{-1}x\left(\frac{y}{1+x^2}\right) \)
\( = \left(\frac{1}{1+x^2}\right)y + y(\tan^{-1}x)^2\left(\frac{1}{1+x^2}\right) - \tan^{-1}x\left(\frac{y}{1+x^2}\right) \)
\( = 0 \)

Therefore, \( y = ce^{\tan^{-1}x} \) satisfies the given differential equation.
In simple words: Find the first and second derivatives of the given function, then substitute them into the differential equation. If both sides equal zero, the function is indeed a solution.

Exam Tip: Carefully apply the chain rule and product rule when differentiating exponential and inverse trigonometric functions - a small error in differentiation will cause the verification to fail.

Question 11. Verify that \( y = ce^{\tan^{-1}x} \) is a solution of the differential equation \( (1 + x^2) \frac{d^2y}{dx^2} + (2x - 1) \frac{dy}{dx} = 0 \)
Answer: Starting with \( y = ce^{\tan^{-1}x} \), we find the derivatives and substitute them into the equation. After differentiation and simplification, the expression evaluates to zero, confirming that the function satisfies the differential equation.
In simple words: Differentiate the function twice and plug these derivatives into the equation. When everything simplifies to zero, you know the function works as a solution.

Exam Tip: Always verify your algebra carefully - collecting like terms and factoring correctly are essential to reach the final zero result.

Question 12. Verify that \( y = ae^{bx} \) is a solution of the differential equation \( \frac{d^2y}{dx^2} = \frac{1}{y}\left(\frac{dy}{dx}\right)^2 \)
Answer: Given \( y = ae^{bx} \), the first derivative is \( \frac{dy}{dx} = abe^{bx} \). The second derivative is \( \frac{d^2y}{dx^2} = ab^2e^{bx} \).

Now we evaluate the right-hand side: \( \frac{1}{y}\left(\frac{dy}{dx}\right)^2 = \frac{1}{ae^{bx}}(abe^{bx})^2 = \frac{a^2b^2e^{2bx}}{ae^{bx}} = ab^2e^{bx} \)

Since \( \frac{d^2y}{dx^2} = ab^2e^{bx} \) and \( \frac{1}{y}\left(\frac{dy}{dx}\right)^2 = ab^2e^{bx} \), both sides are equal. Therefore, \( y = ae^{bx} \) satisfies the differential equation.
In simple words: Compute the first and second derivatives, then verify that the second derivative matches the expression on the right side of the equation.

Exam Tip: When dealing with exponential functions, remember that the derivative of \( e^{bx} \) is always \( be^{bx} \) - this pattern repeats with each differentiation.

Question 13. Verify that \( y = e^{-x} + Ax + B \) is a solution of the differential equation \( e^x\left(\frac{d^2y}{dx^2}\right) = 1 \)
Answer: Starting with \( y = e^{-x} + Ax + B \), we obtain the first derivative: \( \frac{dy}{dx} = -e^{-x} + A \). Taking the second derivative: \( \frac{d^2y}{dx^2} = e^{-x} \).

Now evaluating \( e^x\left(\frac{d^2y}{dx^2}\right) = e^x(e^{-x}) = 1 \).

Since the left side equals 1, which matches the right side, \( y = e^{-x} + Ax + B \) is confirmed as a solution of the given differential equation.
In simple words: Differentiate twice, multiply by \( e^x \), and check if the result equals 1. When it does, the function is a valid solution.

Exam Tip: Note that the arbitrary constants A and B vanish after taking the second derivative - they only affect the first derivative, not the second.

Question 14. Verify that \( Ax^2 + By^2 = 1 \) is a solution of the differential equation \( x\left[y\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2\right] = y\frac{dy}{dx} \)
Answer: Given \( Ax^2 + By^2 = 1 \). Differentiating with respect to x: \( 2Ax + 2By\frac{dy}{dx} = 0 \), which gives \( \frac{dy}{dx} = -\frac{Ax}{By} \).

Differentiating again: \( 2A + 2B\left(\frac{dy}{dx}\right)^2 + 2By\left(\frac{d^2y}{dx^2}\right) = 0 \).

This simplifies to: \( y\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 = -\frac{A}{B} \).

Substituting into the left side of the original equation:
\( x\left[-\frac{A}{B}\right] = -\frac{Ax}{B} \)

And the right side: \( y\left(-\frac{Ax}{By}\right) = -\frac{Ax}{B} \)

Since both sides are equal, \( Ax^2 + By^2 = 1 \) is a valid solution of the differential equation.
In simple words: Use implicit differentiation to find the first and second derivatives, then substitute them into the equation to verify both sides match.

Exam Tip: When working with implicit functions like this, carefully track the chain rule through both differentiation steps to avoid algebraic errors.

Question 15. Verify that \( y = \frac{c - x}{1 + cx} \) is a solution of the differential equation \( (1 + x^2)\frac{dy}{dx} + (1 + y^2) = 0 \)
Answer: Starting with \( y = \frac{c - x}{1 + cx} \). Taking the derivative using the quotient rule:
\( \frac{dy}{dx} = \frac{-1(1 + cx) - (c - x)(c)}{(1 + cx)^2} = \frac{-1 - cx - c^2 + cx}{(1 + cx)^2} = \frac{-(1 + c^2)}{(1 + cx)^2} \)

Now we evaluate \( (1 + x^2)\frac{dy}{dx} + (1 + y^2) \):
\( = (1 + x^2)\left(\frac{-(1 + c^2)}{(1 + cx)^2}\right) + 1 + \left(\frac{c - x}{1 + cx}\right)^2 \)
\( = -\frac{(1 + x^2)(1 + c^2)}{(1 + cx)^2} + \frac{(1 + cx)^2 + (c - x)^2}{(1 + cx)^2} \)

Expanding the numerator: \( (1 + cx)^2 + (c - x)^2 = 1 + 2cx + c^2x^2 + c^2 - 2cx + x^2 = (1 + x^2)(1 + c^2) \)

Therefore, the expression equals \( -\frac{(1 + x^2)(1 + c^2)}{(1 + cx)^2} + \frac{(1 + x^2)(1 + c^2)}{(1 + cx)^2} = 0 \), confirming the solution.
In simple words: Differentiate using the quotient rule, substitute into the equation, and simplify the resulting fraction to verify it equals zero.

Exam Tip: The key is careful algebraic expansion when combining fractions with a common denominator - verify that your numerator simplifies to zero.

Question 16. Verify that \( y = \log\left(x + \sqrt{x^2 + a^2}\right) \) satisfies the differential equation \( \frac{d^2y}{dx^2} + x\frac{dy}{dx} = 0 \)
Answer: Given \( y = \log\left(x + \sqrt{x^2 + a^2}\right) \). The first derivative is:
\( \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \left(1 + \frac{x}{\sqrt{x^2 + a^2}}\right) = \frac{1}{\sqrt{x^2 + a^2}} \)

The second derivative is:
\( \frac{d^2y}{dx^2} = -\frac{x}{(x^2 + a^2)^{3/2}} \)

Now evaluating \( \frac{d^2y}{dx^2} + x\frac{dy}{dx} \):
\( = -\frac{x}{(x^2 + a^2)^{3/2}} + x \cdot \frac{1}{\sqrt{x^2 + a^2}} \)
\( = -\frac{x}{(x^2 + a^2)^{3/2}} + \frac{x}{\sqrt{x^2 + a^2}} \)
\( = -\frac{x}{(x^2 + a^2)^{3/2}} + \frac{x(x^2 + a^2)}{(x^2 + a^2)^{3/2}} = 0 \)

The function satisfies the differential equation.
In simple words: Find the first derivative using the chain rule, then differentiate again. Substitute both derivatives into the equation to confirm they cancel and yield zero.

Exam Tip: When differentiating logarithmic expressions, use the chain rule carefully and simplify complex fractional derivatives by finding a common denominator before substituting.

Question 17. Verify that \( y = e^{-3x} \) is a solution of the differential equation \( \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0 \)
Answer: Starting with \( y = e^{-3x} \). The first derivative is \( \frac{dy}{dx} = -3e^{-3x} \). The second derivative is \( \frac{d^2y}{dx^2} = 9e^{-3x} \).

Substituting into the differential equation:
\( 9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} = 9e^{-3x} - 3e^{-3x} - 6e^{-3x} = 0 \)

Since all terms combine to give zero, \( y = e^{-3x} \) is confirmed as a solution of the equation.
In simple words: Find both derivatives by repeatedly applying the chain rule, then substitute them into the original equation. If the result is zero, the function is a valid solution.

Exam Tip: For exponential functions of the form \( e^{kx} \), each derivative multiplies the function by the constant k. Keep track of the coefficient as you differentiate multiple times.

Exercise 18C

Question 1. Form the differential equation of the family of straight lines \( y = mx + c \), where m and c are arbitrary constants.
Answer: The equation represents a straight line as \( y = mx + c \). Differentiating with respect to x:
\( \frac{dy}{dx} = m \)

Differentiating once more with respect to x:
\( \frac{d^2y}{dx^2} = 0 \)

This is the differential equation of the family of straight lines \( y = mx + c \), where m and c are arbitrary constants.
In simple words: Since a straight line has constant slope, its second derivative must be zero. This simple condition captures all lines with any slope and any y-intercept.

Exam Tip: The differential equation of a family is free of arbitrary constants - you eliminate them by successive differentiation until they disappear.

Question 2. Form the differential equation of the family of concentric circles \( x^2 + y^2 = a^2 \), where \( a > 0 \) and a is a parameter.
Answer: For the family of concentric circles \( x^2 + y^2 = a^2 \) where a represents the radius and is an arbitrary constant, the circles are all centered at the origin. Differentiating with respect to x:
\( 2x + 2y\frac{dy}{dx} = 0 \)

Simplifying:
\( x + y\frac{dy}{dx} = 0 \)

This is the required differential equation of the family of concentric circles.
In simple words: Differentiate the circle equation implicitly with respect to x. The result shows how x and y are related through their rates of change at any point on the circle.

Exam Tip: When differentiating implicit equations, remember that y is a function of x - apply the chain rule to terms involving y.

Question 3. Form the differential equation of the family of curves \( y = a\sin(bx + c) \), where a and c are parameters.
Answer: Given the family \( y = a\sin(bx + c) \) where a and c are parameters. Differentiating with respect to x:
\( \frac{dy}{dx} = ab\cos(bx + c) \) ... (1)

Differentiating again:
\( \frac{d^2y}{dx^2} = -ab^2\sin(bx + c) = -b^2y \) (substituting the original equation)

Therefore:
\( \frac{d^2y}{dx^2} + b^2y = 0 \)

This is the required differential equation of the family.
In simple words: After two differentiations, the original function reappears multiplied by \( -b^2 \). Rearranging gives the final differential equation without any arbitrary constants.

Exam Tip: For trigonometric families, successive differentiation will eventually reproduce the original function with a coefficient - use this to eliminate the parameters.

Question 4. Form the differential equation of the family of curves \( x = A\cos nt + B\sin nt \), where A and B are arbitrary constants.
Answer: For the family of curves \( x = A\cos nt + B\sin nt \) where A and B are arbitrary constants. Differentiating with respect to t:
\( \frac{dx}{dt} = -An\sin nt + Bn\cos nt \) ... (1)

Differentiating again:
\( \frac{d^2x}{dt^2} = -An^2\cos nt - Bn^2\sin nt = -n^2(A\cos nt + B\sin nt) = -n^2 x \) (substituting the original)

Therefore:
\( \frac{d^2x}{dt^2} + n^2x = 0 \)

This is the required differential equation of the family.
In simple words: Differentiate twice, and the second derivative reproduces the original expression scaled by \( -n^2 \). Rearranging yields the differential equation free of A and B.

Exam Tip: This type of equation describes simple harmonic motion - the second derivative is proportional to the negative of the function itself.

Question 5. Form the differential equation of the family of curves \( y = ae^{bx} \), where a and b are arbitrary constants.
Answer: Given the family \( y = ae^{bx} \) where a and b are arbitrary constants. Differentiating with respect to x:
\( \frac{dy}{dx} = abe^{bx} \) ... (1)

Differentiating again:
\( \frac{d^2y}{dx^2} = ab^2e^{bx} \) ... (2)

From equation (1): \( ab e^{bx} = \frac{dy}{dx} \)

Substituting into equation (2):
\( \frac{d^2y}{dx^2} = b \cdot \frac{dy}{dx} \)

Or alternatively, multiplying both sides by y:
\( y\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \)

This is the required differential equation of the family.
In simple words: From the first derivative, extract the coefficient b. Show that this b relates the second derivative to the first derivative through the original function.

Exam Tip: When you have two unknown parameters, you need two differentiations to eliminate both - the second equation should contain no a or b terms.

Question 6. Form the differential equation of the family of curves \( y^2 = m(a^2 - x^2) \), where a and m are parameters.
Answer: Given the family \( y^2 = m(a^2 - x^2) \) where a and m are parameters. Differentiating with respect to x:
\( 2y\frac{dy}{dx} = -2mx \)
\( y\frac{dy}{dx} = -mx \) ... (1)

Differentiating again:
\( \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = -m \) ... (2)

From equation (1): \( m = -\frac{y}{x}\frac{dy}{dx} \)

Substituting into equation (2):
\( \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = \frac{y}{x}\frac{dy}{dx} \)

Simplifying:
\( xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0 \)

This is the required differential equation.
In simple words: Use implicit differentiation twice to get two equations. From the first, solve for one parameter and substitute into the second to eliminate both parameters.

Exam Tip: When dealing with multiple parameters, organize your working clearly to track which parameter you are eliminating at each step.

Question 7. Form the differential equation of the family of curves given by \( (x - a)^2 + 2y^2 = a^2 \), where a is an arbitrary constant.
Answer: For the family of curves \( (x - a)^2 + 2y^2 = a^2 \) where a is an arbitrary constant. Expanding:
\( x^2 - 2ax + a^2 + 2y^2 = a^2 \)
\( x^2 - 2ax + 2y^2 = 0 \) ... (1)

Differentiating with respect to x:
\( 2x - 2a + 4y\frac{dy}{dx} = 0 \)
\( x - a + 2y\frac{dy}{dx} = 0 \)
\( a = x + 2y\frac{dy}{dx} \) ... (2)

Substituting (2) into (1):
\( x^2 - 2\left(x + 2y\frac{dy}{dx}\right)x + 2y^2 = 0 \)
\( x^2 - 2x^2 - 4xy\frac{dy}{dx} + 2y^2 = 0 \)
\( -x^2 - 4xy\frac{dy}{dx} + 2y^2 = 0 \)
\( 4xy\frac{dy}{dx} = 2y^2 - x^2 \)
\( \frac{dy}{dx} = \frac{2y^2 - x^2}{4xy} \)

This is the required differential equation.
In simple words: First simplify the original equation, then differentiate to find an expression for the parameter a. Substitute back to get an equation involving only x, y, and dy/dx.

Exam Tip: Always expand and simplify the original equation before differentiating - this often makes the algebra cleaner and reduces computational errors.

Question 8. Form the differential equation of the family of curves given by \( x^2 + y^2 - 2ay = a^2 \), where a is an arbitrary constant.
Answer: For the family of curves \( x^2 + y^2 - 2ay = a^2 \) where a is an arbitrary constant. Rearranging:
\( x^2 + y^2 - 2ay - a^2 = 0 \)
\( x^2 + (y - a)^2 = 2a^2 \) ... (1)

Differentiating with respect to x:
\( 2x + 2(y - a)\frac{dy}{dx} = 0 \)
\( x + (y - a)\frac{dy}{dx} = 0 \)
\( y - a = -\frac{x}{\frac{dy}{dx}} \) ... (2)

Differentiating again:
\( 2x + 2\left(\frac{dy}{dx}\right)^2 + 2(y - a)\frac{d^2y}{dx^2} = 0 \)
\( x + \left(\frac{dy}{dx}\right)^2 + (y - a)\frac{d^2y}{dx^2} = 0 \)
\( x + \left(\frac{dy}{dx}\right)^2 - \frac{x}{\frac{dy}{dx}} \cdot \frac{d^2y}{dx^2} = 0 \) (using equation 2)

Simplifying:
\( x\left(\frac{dy}{dx}\right)^2 + \left(\frac{dy}{dx}\right)^3 - x\frac{d^2y}{dx^2} = 0 \)

Or equivalently:\
\( \left[\left(\frac{dy}{dx}\right)^2 + 1\right]x\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \)

This is the required differential equation.
In simple words: Differentiate the original equation twice to eliminate the parameter a. The first derivative gives you an expression for a, which you substitute into the second derivative equation.

Exam Tip: Keep track of your substitutions carefully - it helps to mark key relationships (like what a equals) so you don't lose track during complex algebraic manipulations.

Question 9. Form the differential equation of the family of all circles touching the y-axis at the origin.
Answer: A circle that touches the y-axis at the origin must have its center on the x-axis. If the radius is a (which equals the x-coordinate of the center), the equation becomes:
\( (x - a)^2 + y^2 = a^2 \)

Expanding:
\( x^2 - 2ax + a^2 + y^2 = a^2 \)
\( x^2 + y^2 = 2ax \) ... (1)

Differentiating with respect to x:
\( 2x + 2y\frac{dy}{dx} = 2a \)
\( x + y\frac{dy}{dx} = a \)
\( a = x + y\frac{dy}{dx} \) ... (2)

Substituting (2) into (1):
\( x^2 + y^2 = 2x\left(x + y\frac{dy}{dx}\right) \)
\( x^2 + y^2 = 2x^2 + 2xy\frac{dy}{dx} \)
\( y^2 - x^2 = 2xy\frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \)

This is the required differential equation of all circles touching the y-axis at the origin.
In simple words: Use the geometric constraint (center on x-axis, touching at origin) to determine the family equation. Then differentiate and eliminate the parameter to find the differential equation.

Exam Tip: When a geometric condition is given, translate it carefully into an algebraic equation first before proceeding with the standard method of finding the differential equation.

Question 10. Form the differential equation of the family of circles having centers on the y-axis and radius 2 units.
Answer: Circles with centers on the y-axis and radius 2 have the form:
\( x^2 + (y - a)^2 = 4 \)

where a is an arbitrary constant representing the y-coordinate of the center. Expanding:
\( x^2 + y^2 - 2ay + a^2 = 4 \) ... (1)

Differentiating with respect to x:
\( 2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0 \)
\( x + y\frac{dy}{dx} - a\frac{dy}{dx} = 0 \)
\( a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \) ... (2)

Differentiating (2):
\( 2x + 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} = 0 \)
\( x + \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} = 0 \)

Rearranging:
\( x^2 + \left(\frac{x}{\frac{dy}{dx}}\right)^2 = 4 \)
\( x^2 + \frac{x^2}{\left(\frac{dy}{dx}\right)^2} = 4 \)
\( x^2\left[1 + \frac{1}{\left(\frac{dy}{dx}\right)^2}\right] = 4 \)
\( x^2\left[\frac{\left(\frac{dy}{dx}\right)^2 + 1}{\left(\frac{dy}{dx}\right)^2}\right] = 4 \)

This is the required differential equation.
In simple words: Since all circles have the same radius and centers on the y-axis, only the y-coordinate varies. Differentiate to eliminate this parameter and get the differential equation.

Exam Tip: For families with a fixed property (like constant radius), that constraint simplifies the general form and reduces the number of arbitrary constants to eliminate.

Question 11. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
Answer: Circles in the second quadrant touching both coordinate axes must have centers at \( (-a, a) \) where \( a > 0 \), with radius a. The equation is:
\( (x + a)^2 + (y - a)^2 = a^2 \)

Expanding:
\( x^2 + 2ax + a^2 + y^2 - 2ay + a^2 = a^2 \)
\( x^2 + y^2 + 2ax - 2ay + a^2 = 0 \) ... (1)

Differentiating with respect to x:
\( 2x + 2y\frac{dy}{dx} + 2a - 2a\frac{dy}{dx} = 0 \)
\( x + y\frac{dy}{dx} + a - a\frac{dy}{dx} = 0 \)
\( a = \frac{-(x + y\frac{dy}{dx})}{1 - \frac{dy}{dx}} = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx} - 1} \) ... (2)

Substituting (2) into (1) and simplifying (after considerable algebra):
\( (x + y)^2\left[\left(\frac{dy}{dx}\right)^2 + 1\right] = \left(x + y\frac{dy}{dx}\right)^2 \)

This is the required differential equation.
In simple words: The geometric constraint (second quadrant, touching both axes) determines that centers lie on the line y = -x. Use this to write the family equation, then differentiate to eliminate the parameter.

Exam Tip: When circles touch coordinate axes, use symmetry arguments: if a circle in the second quadrant touches both axes, its center must be equidistant from both axes.

Question 12. Form the differential equation of the family of circles having centers on the x-axis and radius unity.
Answer: Circles with centers on the x-axis and radius 1 have the form:
\( (x - a)^2 + y^2 = 1 \)

where a is an arbitrary constant representing the x-coordinate of the center. Expanding:
\( x^2 - 2ax + a^2 + y^2 = 1 \) ... (1)

Differentiating with respect to x:
\( 2x - 2a + 2y\frac{dy}{dx} = 0 \)
\( x - a + y\frac{dy}{dx} = 0 \)
\( a = x + y\frac{dy}{dx} \) ... (2)

Substituting (2) into (1):
\( x^2 - 2x(x + y\frac{dy}{dx}) + (x + y\frac{dy}{dx})^2 + y^2 = 1 \)
\( x^2 - 2x^2 - 2xy\frac{dy}{dx} + x^2 + 2xy\frac{dy}{dx} + y^2\left(\frac{dy}{dx}\right)^2 + y^2 = 1 \)
\( y^2\left(\frac{dy}{dx}\right)^2 + y^2 = 1 \)
\( y^2\left[\left(\frac{dy}{dx}\right)^2 + 1\right] = 1 \)

This is the required differential equation.
In simple words: The center can slide along the x-axis while the radius stays fixed at 1. Differentiate the circle equation to eliminate the parameter a and obtain the differential equation.

Exam Tip: When the radius is constant, the parameter to eliminate is just the position of the center. One differentiation is usually sufficient.

Question 13. Form the differential equation of the family of circles passing through the fixed point \( (a, 0) \) and \( (-a, 0) \), where a is the parameter.
Answer: Circles passing through both \( (a, 0) \) and \( (-a, 0) \) have their centers on the y-axis (by symmetry). Let the center be \( (0, h) \) where h is an arbitrary constant. The radius is found using the distance formula:
\( r = \sqrt{a^2 + h^2} \)

The equation is:
\( x^2 + (y - h)^2 = a^2 + h^2 \)
\( x^2 + y^2 - 2yh + h^2 = a^2 + h^2 \)
\( x^2 + y^2 = a^2 + 2yh \) ... (1)

Differentiating with respect to x:
\( 2x + 2y\frac{dy}{dx} = 2h\frac{dy}{dx} \)
\( x + y\frac{dy}{dx} = h\frac{dy}{dx} \)
\( h = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} = \frac{x}{\frac{dy}{dx}} + y \) ... (2)

Substituting (2) into (1):
\( x^2 + y^2 = a^2 + 2y\left(\frac{x}{\frac{dy}{dx}} + y\right) \)
\( x^2 + y^2 = a^2 + \frac{2xy}{\frac{dy}{dx}} + 2y^2 \)
\( x^2 - y^2 - a^2 = \frac{2xy}{\frac{dy}{dx}} \)
\( \left(\frac{dy}{dx}\right)(x^2 - y^2 - a^2) = 2xy \)

This is the required differential equation of the family.
In simple words: Use the symmetry of the two fixed points to determine that centers lie on the perpendicular bisector (the y-axis). Then express the family equation and eliminate the parameter h through differentiation.

Exam Tip: Symmetry is a powerful tool - when two points have the same y-coordinate and opposite x-coordinates, any circle through them has its center on the y-axis.

Question 14. Form the differential equation of the family of parabolas having a vertex at the origin and axis along positive y-axis.
Answer: The parabola family with its vertex centered at the origin and axis running along the positive y-direction has the standard form \( x^2 = 4ay \), where \( a \) is an arbitrary constant.

Starting with \( x^2 = 4ay \) ... (1)

Taking the derivative with respect to \( x \) on both sides:
\[ 2x = 4a\frac{dy}{dx} \]

From this, we obtain:
\[ a = \frac{x}{2\frac{dy}{dx}} \]

Now substitute this expression for \( a \) back into equation (1):
\[ x^2 = 4 \cdot \frac{x}{2\frac{dy}{dx}} \cdot y \]

Simplifying:
\[ x\frac{dy}{dx} = 2y \]

This gives us the required differential equation. In simple words: Start with the standard parabola equation, take its derivative to find the slope, eliminate the constant by substitution, and you get a simple relationship between x, y, and the slope dy/dx.

Exam Tip: Always differentiate the original family equation first, then use the derivative to find the arbitrary constant in terms of x and y before substituting back into the family equation.

Question 15. Form the differential equation of the family of an ellipse having foci on the y-axis and centers at the origin.
Answer: An ellipse with its foci positioned on the y-axis and its center at the origin takes the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) ... (1), where \( a \) and \( b \) are arbitrary constants.

Differentiating equation (1) with respect to \( x \):
\[ \frac{2x}{b^2} + \frac{2y}{a^2}\frac{dy}{dx} = 0 \]

Simplifying:
\[ \frac{x}{b^2} + \frac{y}{a^2}\frac{dy}{dx} = 0 \]

This gives:
\[ \frac{y}{a^2}\frac{dy}{dx} = -\frac{x}{b^2} \]

Or:
\[ \frac{y}{x}\frac{dy}{dx} = -\frac{a^2}{b^2} \] ... (2)

Differentiating equation (2) again with respect to \( x \):
\[ \frac{d}{dx}\left(\frac{y}{x}\frac{dy}{dx}\right) = 0 \]

Expanding using the product rule:
\[ \frac{y}{x}\frac{d^2y}{dx^2} + \frac{dy}{dx}\left(\frac{d}{dx}\left(\frac{y}{x}\right)\right) = 0 \]

Since \( \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y}{x^2} \):
\[ \frac{y}{x}\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x\frac{dy}{dx} - y}{x^2} = 0 \]

Rearranging:
\[ (x^2 - a^2 - y^2)\frac{dy}{dx} = 2xy \]

This is the required differential equation. In simple words: Differentiate the ellipse equation twice, use the chain rule to eliminate the constants, and combine the results to get a single differential equation connecting x, y, and their derivatives.

Exam Tip: When eliminating two constants, you need to differentiate twice. After each differentiation, check that you are progressively removing one constant until none remain.

Question 16. Form the differential equation of the family of hyperbolas having foci on the x-axis and centers at the origin.
Answer: A hyperbola with foci positioned on the x-axis and center at the origin has the standard equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) ... (1), where \( a \) and \( b \) are arbitrary constants.

Differentiating equation (1) with respect to \( x \):
\[ \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0 \]

Simplifying:
\[ \frac{x}{a^2} - \frac{y}{b^2}\frac{dy}{dx} = 0 \]

Rearranging:
\[ \frac{y}{b^2}\frac{dy}{dx} = \frac{x}{a^2} \]

Or:
\[ \frac{y}{x}\frac{dy}{dx} = \frac{b^2}{a^2} \] ... (2)

Differentiating equation (2) again with respect to \( x \):
\[ \frac{d}{dx}\left(\frac{y}{x}\frac{dy}{dx}\right) = 0 \]

Expanding using the product rule:
\[ \frac{y}{x}\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) = 0 \]

Since \( \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x\frac{dy}{dx} - y}{x^2} \):
\[ \frac{y}{x}\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x\frac{dy}{dx} - y}{x^2} = 0 \]

Rearranging:
\[ xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0 \]

This is the required differential equation. In simple words: Apply differentiation twice to the hyperbola equation, use algebraic techniques to remove the arbitrary constants, and simplify to arrive at a pure differential equation without any unknown parameters.

Exam Tip: For conic sections, recognizing the standard forms and the number of arbitrary constants helps determine how many times you need to differentiate. Two constants require two differentiations.