CBSE Class 12 Mathematics Differential Equations Worksheet Set 01

CBSE Class 12 Mathematics Differentials Equation (1). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

1 MARK QUESTIONS

Question. What is the degree of the following differential equation?
\( 5x\left(\frac{dy}{dx}\right)^2 - \frac{d^2y}{dx^2} - 6y = \log x \)
Answer: The highest order derivative in the given differential equation is \( \frac{d^2y}{dx^2} \), which has an order of 2. The exponent of this highest order derivative is 1. Since the differential equation is a polynomial in its derivatives, the degree of the differential equation is 1.

Question. Write the degree of the differential equation
\( x^3 \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^4 = 0 \)
Answer: The highest order derivative in the given differential equation is \( \frac{d^2y}{dx^2} \), and its power is 2. Therefore, the degree of the differential equation is 2.

Question. Determine the order and degree of \( t^2 \frac{d^2s}{dt^2} - st \frac{ds}{dt} = s \). And, also state if it is linear or non linear.
Answer: For the given differential equation \( t^2 \frac{d^2s}{dt^2} - st \frac{ds}{dt} = s \):
- Order: The highest derivative present is \( \frac{d^2s}{dt^2} \), so the order is 2.
- Degree: The power of the highest order derivative \( \frac{d^2s}{dt^2} \) is 1, so the degree is 1.
- Linearity: The equation is non-linear because it contains the product of the dependent variable \( s \) and its derivative \( \frac{ds}{dt} \) in the term \( st \frac{ds}{dt} \).

Question. Determine the order and degree of the differential equation:
\( y = px + \sqrt{a^2p^2 + b^2} \), where \( p = \frac{dy}{dx} \)
Answer: First, we write the differential equation in a polynomial form with respect to its derivatives:
\( y - px = \sqrt{a^2p^2 + b^2} \)
Squaring both sides:
\( (y - px)^2 = a^2p^2 + b^2 \)
\( \Rightarrow y^2 - 2xyp + x^2p^2 = a^2p^2 + b^2 \)
Substituting \( p = \frac{dy}{dx} \):
\( y^2 - 2xy \left(\frac{dy}{dx}\right) + (x^2 - a^2)\left(\frac{dy}{dx}\right)^2 - b^2 = 0 \)
- Order: The highest order derivative present is \( \frac{dy}{dx} \), so the order is 1.
- Degree: The power of \( \frac{dy}{dx} \) in this polynomial form is 2, so the degree is 2.

Question. Find the integrating factor for the following differential equation:
\( x \log x \frac{dy}{dx} + y = 2\log x \)
Answer: Dividing the entire equation by \( x \log x \) to express it in standard linear form \( \frac{dy}{dx} + Py = Q \):
\( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x} \)
Here, \( P = \frac{1}{x \log x} \).
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \frac{1}{x \log x} \, dx} \)
To evaluate \( \int \frac{1}{x \log x} \, dx \), let \( \log x = u \implies \frac{1}{x} \, dx = du \):
\( \int \frac{1}{u} \, du = \log|u| = \log|\log x| \)
Thus, \( \text{I.F.} = e^{\log(\log x)} = \log x \).

Question. Find the differential equation of the family of lines passing through the origin
Answer: The equation of the family of lines passing through the origin is:
\( y = mx \) --- (1)
where \( m \) is an arbitrary constant. Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = m \)
Substituting the value of \( m \) from (1) into this derivative equation:
\( \frac{dy}{dx} = \frac{y}{x} \)
\( \Rightarrow x \frac{dy}{dx} - y = 0 \)

Question. Find the differential equation of all circles, which pass through the origin and whose centres lie on the Y-axis.
Answer: Let the center of the circle on the Y-axis be \( (0, b) \). Since the circle passes through the origin \( (0,0) \), its radius is \( |b| \).
The equation of such a circle is:
\( (x - 0)^2 + (y - b)^2 = b^2 \)
\( \Rightarrow x^2 + y^2 - 2by + b^2 = b^2 \)
\( \Rightarrow x^2 + y^2 = 2by \)
\( \Rightarrow 2b = \frac{x^2 + y^2}{y} \) --- (1)
Differentiating \( x^2 + y^2 = 2by \) with respect to \( x \):
\( 2x + 2y \frac{dy}{dx} = 2b \frac{dy}{dx} \)
Substitute \( 2b \) from (1):
\( 2x + 2y \frac{dy}{dx} = \left(\frac{x^2 + y^2}{y}\right) \frac{dy}{dx} \)
\( \Rightarrow 2xy + 2y^2 \frac{dy}{dx} = (x^2 + y^2) \frac{dy}{dx} \)
\( \Rightarrow (x^2 - y^2) \frac{dy}{dx} - 2xy = 0 \)

Question. Show that the differential equation of which \( y = 2(x^2 - 1) + ce^{-x^2} \) is a solution to \( \frac{dy}{dx} + 2xy = 4x^3 \).
Answer: Given function:
\( y = 2(x^2 - 1) + ce^{-x^2} = 2x^2 - 2 + ce^{-x^2} \)
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = 4x + ce^{-x^2}(-2x) \)
\( \Rightarrow \frac{dy}{dx} = 4x - 2x(ce^{-x^2}) \)
From the given equation, \( ce^{-x^2} = y - 2x^2 + 2 \). Substituting this:
\( \frac{dy}{dx} = 4x - 2x(y - 2x^2 + 2) \)
\( \Rightarrow \frac{dy}{dx} = 4x - 2xy + 4x^3 - 4x \)
\( \Rightarrow \frac{dy}{dx} + 2xy = 4x^3 \)
Since LHS = RHS, the given function is a solution to the differential equation.

Question. If \( y\sqrt{x^2 + 1} = \log\left[\sqrt{x^2 + 1} - x\right] \), Show that \( (x^2 + 1)\frac{dy}{dx} + xy + 1 = 0 \)
Answer: Given:
\( y\sqrt{x^2 + 1} = \log\left[\sqrt{x^2 + 1} - x\right] \)
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx}\sqrt{x^2 + 1} + y \left( \frac{2x}{2\sqrt{x^2 + 1}} \right) = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{2x}{2\sqrt{x^2 + 1}} - 1 \right) \)
\( \Rightarrow \frac{dy}{dx}\sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \left( \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right) \)
\( \Rightarrow \frac{dy}{dx}\sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \frac{-(\sqrt{x^2 + 1} - x)}{\sqrt{x^2 + 1}} \)
\( \Rightarrow \frac{dy}{dx}\sqrt{x^2 + 1} + \frac{xy}{\sqrt{x^2 + 1}} = -\frac{1}{\sqrt{x^2 + 1}} \)
Multiplying the entire equation by \( \sqrt{x^2 + 1} \):
\( (x^2 + 1)\frac{dy}{dx} + xy = -1 \)
\( \Rightarrow (x^2 + 1)\frac{dy}{dx} + xy + 1 = 0 \)
Hence proved.

Question. Form the differential equation of the family of curves represented by the equation:
\( (2x + a)^2 + y^2 = a^2 \).
Answer: Given equation:
\( (2x + a)^2 + y^2 = a^2 \)
\( \Rightarrow 4x^2 + 4ax + a^2 + y^2 = a^2 \)
\( \Rightarrow 4x^2 + 4ax + y^2 = 0 \)
\( \Rightarrow 4a = -\frac{4x^2 + y^2}{x} \) --- (1)
Differentiating \( 4x^2 + 4ax + y^2 = 0 \) with respect to \( x \):
\( 8x + 4a + 2y\frac{dy}{dx} = 0 \)
Substituting \( 4a \) from (1):
\( 8x - \left(\frac{4x^2 + y^2}{x}\right) + 2y\frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{8x^2 - 4x^2 - y^2 + 2xy\frac{dy}{dx}}{x} = 0 \)
\( \Rightarrow 4x^2 - y^2 + 2xy\frac{dy}{dx} = 0 \)
\( \Rightarrow (y^2 - 4x^2) - 2xy\frac{dy}{dx} = 0 \)

Question. Write the degree of the differential equation
\( x^3\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^4 = 0 \)
Answer: The highest order derivative in this differential equation is \( \frac{d^2y}{dx^2} \). The power raised to this highest derivative is 2. Therefore, the degree of this differential equation is 2.

Question. Solve the following differential equation:
\( \cos^2 x \frac{dy}{dx} + y = \tan x \)
Answer: Dividing the entire equation by \( \cos^2 x \):
\( \frac{dy}{dx} + (\sec^2 x) y = \tan x \sec^2 x \)
This is a linear differential of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \sec^2 x \) and \( Q = \tan x \sec^2 x \).
Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \sec^2 x \, dx} = e^{\tan x} \)
The general solution is given by:
\( y \cdot \text{I.F.} = \int (Q \cdot \text{I.F.}) \, dx + C \)
\( \Rightarrow y \cdot e^{\tan x} = \int e^{\tan x} \tan x \sec^2 x \, dx + C \)
To solve the integral, let \( \tan x = t \implies \sec^2 x \, dx = dt \):
\( \int e^t \cdot t \, dt = t e^t - \int e^t \, dt = t e^t - e^t = e^t(t - 1) = e^{\tan x}(\tan x - 1) \)
Thus:
\( y e^{\tan x} = e^{\tan x}(\tan x - 1) + C \)
\( \Rightarrow y = \tan x - 1 + C e^{-\tan x} \)

4 MARKS OR 6 MARKS QUESTIONS

Question. Solve the following differential equation:
\( (x^2 - y^2) \, dx + 2xy \, dy = 0 \) given that \( y = 1 \) when \( x = 1 \)
Answer: Rearranging the given equation:
\( 2xy \, dy = -(x^2 - y^2) \, dx \)
\( \Rightarrow \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \) --- (1)
This is a homogeneous differential equation. Let \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting into (1):
\( v + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{v^2x^2 - x^2}{2vx^2} = \frac{v^2 - 1}{2v} \)
\( \Rightarrow x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\left(\frac{v^2 + 1}{2v}\right) \)
Separating the variables:
\( \frac{2v}{v^2 + 1} \, dv = -\frac{1}{x} \, dx \)
Integrating both sides:
\( \int \frac{2v}{v^2 + 1} \, dv = -\int \frac{1}{x} \, dx \)
\( \Rightarrow \log(v^2 + 1) = -\log|x| + \log C \)
\( \Rightarrow \log(v^2 + 1) = \log\left(\frac{C}{x}\right) \)
\( \Rightarrow v^2 + 1 = \frac{C}{x} \)
Substitute \( v = \frac{y}{x} \):
\( \frac{y^2}{x^2} + 1 = \frac{C}{x} \)
\( \Rightarrow \frac{y^2 + x^2}{x^2} = \frac{C}{x} \)
\( \Rightarrow x^2 + y^2 = Cx \) --- (2)
Using the initial condition \( y = 1 \) when \( x = 1 \):
\( 1^2 + 1^2 = C(1) \implies C = 2 \)
Substituting \( C = 2 \) into (2) gives the particular solution:
\( x^2 + y^2 = 2x \)

Question. Find the particular solution, satisfying the given condition, for the following differential equation:
\( \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \); \( y = 0 \) when \( x = 1 \)
Answer: Let \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting into the differential equation:
\( v + x\frac{dv}{dx} - v + \csc v = 0 \)
\( \Rightarrow x\frac{dv}{dx} + \csc v = 0 \)
\( \Rightarrow x\frac{dv}{dx} = -\csc v = -\frac{1}{\sin v} \)
Separating variables:
\( \sin v \, dv = -\frac{1}{x} \, dx \)
Integrating both sides:
\( \int \sin v \, dv = -\int \frac{1}{x} \, dx \)
\( \Rightarrow -\cos v = -\log|x| + C \)
\( \Rightarrow \cos\left(\frac{y}{x}\right) = \log|x| - C \)
Let \( C' = -C \):
\( \cos\left(\frac{y}{x}\right) = \log|x| + C' \) --- (1)
Using the given condition \( y = 0 \) when \( x = 1 \):
\( \cos(0) = \log(1) + C' \)
\( \Rightarrow 1 = 0 + C' \implies C' = 1 \)
Substituting \( C' = 1 \) into (1), we get the particular solution:
\( \cos\left(\frac{y}{x}\right) = \log|x| + 1 \)

Question. Find the particular solution of the differential equation satisfying the given conditions:
\( x^2 \, dy + (xy + y^2) \, dx = 0 \); \( y = 1 \) when \( x = 1 \).
Answer: Rearranging the given equation:
\( x^2 \, dy = -(xy + y^2) \, dx \)
\( \Rightarrow \frac{dy}{dx} = -\left(\frac{xy + y^2}{x^2}\right) \) --- (1)
This is a homogeneous differential equation. Let \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting into (1):
\( v + x\frac{dv}{dx} = -\left(\frac{v x^2 + v^2 x^2}{x^2}\right) = -(v + v^2) \)
\( \Rightarrow x\frac{dv}{dx} = -v - v^2 - v = -(v^2 + 2v) \)
Separating variables:
\( \frac{dv}{v(v+2)} = -\frac{1}{x} \, dx \)
Integrating both sides:
\( \int \frac{dv}{v(v+2)} = -\int \frac{1}{x} \, dx \)
Using partial fractions, \( \frac{1}{v(v+2)} = \frac{1}{2}\left[\frac{1}{v} - \frac{1}{v+2}\right] \):
\( \frac{1}{2}\int \left(\frac{1}{v} - \frac{1}{v+2}\right) \, dv = -\int \frac{1}{x} \, dx \)
\( \Rightarrow \frac{1}{2}\left(\log|v| - \log|v+2|\right) = -\log|x| + \log C \)
\( \Rightarrow \log\sqrt{\frac{v}{v+2}} = \log\left(\frac{C}{x}\right) \)
\( \Rightarrow \sqrt{\frac{v}{v+2}} = \frac{C}{x} \)
Squaring both sides:
\( \frac{v}{v+2} = \frac{C^2}{x^2} \)
Let \( C' = C^2 \). Substituting \( v = \frac{y}{x} \):
\( \frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C'}{x^2} \)
\( \Rightarrow \frac{y}{y + 2x} = \frac{C'}{x^2} \)
\( \Rightarrow x^2 y = C'(y + 2x) \) --- (2)
Using the condition \( y = 1 \) when \( x = 1 \):
\( 1^2(1) = C'(1 + 2(1)) \implies 1 = 3C' \implies C' = \frac{1}{3} \)
Substituting \( C' = \frac{1}{3} \) into (2):
\( x^2 y = \frac{1}{3}(y + 2x) \)
\( \Rightarrow 3x^2 y = y + 2x \)

Question. Find the general solution of the differential equation,
\( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \)
Answer: Dividing by \( x\log x \) to write the equation in the standard linear form \( \frac{dy}{dx} + Py = Q \):
\( \frac{dy}{dx} + \frac{1}{x\log x} y = \frac{2}{x^2} \)
Here, \( P = \frac{1}{x\log x} \) and \( Q = \frac{2}{x^2} \).
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int \frac{1}{x\log x} \, dx} = e^{\log(\log x)} = \log x \)
The general solution is:
\( y \cdot \text{I.F.} = \int (Q \cdot \text{I.F.}) \, dx + C \)
\( \Rightarrow y \log x = \int \frac{2}{x^2} \log x \, dx + C \)
To find \( \int \frac{\log x}{x^2} \, dx \), we use integration by parts:
Let \( u = \log x \implies du = \frac{1}{x} \, dx \)
Let \( dv = x^{-2} \, dx \implies v = -\frac{1}{x} \)
\( \int \frac{\log x}{x^2} \, dx = -\frac{\log x}{x} - \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}\right) \, dx = -\frac{\log x}{x} + \int x^{-2} \, dx = -\frac{\log x}{x} - \frac{1}{x} = -\frac{\log x + 1}{x} \)
Substituting this back:
\( y \log x = 2\left(-\frac{\log x + 1}{x}\right) + C \)
\( \Rightarrow y \log x = -\frac{2(\log x + 1)}{x} + C \)

Question. Solve the following differential equation:
\( e^x \tan y \, dx + (1 - e^x)\sec^2 y \, dy = 0 \)
Answer: Rearranging the given equation to separate variables:
\( (1 - e^x)\sec^2 y \, dy = -e^x \tan y \, dx \)
\( \Rightarrow \frac{\sec^2 y}{\tan y} \, dy = -\frac{e^x}{1 - e^x} \, dx \)
\( \Rightarrow \frac{\sec^2 y}{\tan y} \, dy = \frac{e^x}{e^x - 1} \, dx \)
Integrating both sides:
\( \int \frac{\sec^2 y}{\tan y} \, dy = \int \frac{e^x}{e^x - 1} \, dx \)
\( \Rightarrow \log|\tan y| = \log|e^x - 1| + \log C \)
\( \Rightarrow \log|\tan y| = \log|C(e^x - 1)| \)
\( \Rightarrow \tan y = C(e^x - 1) \)

Question. Find the particular solution of the following differential equation:
\( (x+1)\frac{dy}{dx} = 2e^{-y} - 1 \); \( y = 0 \) when \( x = 0 \)
Answer: Separating the variables:
\( \frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1} \)
\( \Rightarrow \frac{dy}{\frac{2}{e^y} - 1} = \frac{dx}{x+1} \)
\( \Rightarrow \frac{e^y \, dy}{2 - e^y} = \frac{dx}{x+1} \)
Integrating both sides:
\( \int \frac{e^y}{2 - e^y} \, dy = \int \frac{1}{x+1} \, dx \)
Let \( 2 - e^y = u \implies -e^y \, dy = du \):
\( -\int \frac{1}{u} \, du = \int \frac{1}{x+1} \, dx \)
\( \Rightarrow -\log|2 - e^y| = \log|x+1| + \log C \)
\( \Rightarrow \log|x+1| + \log|2 - e^y| = \log C' \) (where \( \log C' = -\log C \))
\( \Rightarrow (x+1)(2 - e^y) = C' \) --- (1)
Applying the initial condition \( y = 0 \) when \( x = 0 \):
\( (0+1)(2 - e^0) = C' \)
\( \Rightarrow 1(2-1) = C' \implies C' = 1 \)
Substituting \( C' = 1 \) into (1):
\( (x+1)(2 - e^y) = 1 \)
\( \Rightarrow 2 - e^y = \frac{1}{x+1} \)
\( \Rightarrow e^y = 2 - \frac{1}{x+1} = \frac{2x+1}{x+1} \)
\( \Rightarrow y = \log\left(\frac{2x+1}{x+1}\right) \)

Question. Find the particular solution of the differential equation
\( \log\left(\frac{dy}{dx}\right) = 3x + 4y \), given that \( y = 0 \) when \( x = 0 \).
Answer: Rewriting the equation by taking exponentials:
\( \frac{dy}{dx} = e^{3x+4y} = e^{3x} \cdot e^{4y} \)
Separating variables:
\( e^{-4y} \, dy = e^{3x} \, dx \)
Integrating both sides:
\( \int e^{-4y} \, dy = \int e^{3x} \, dx \)
\( \Rightarrow \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \)
Multiplying the entire equation by \( -12 \):
\( 3e^{-4y} = -4e^{3x} + C \)
\( \Rightarrow 4e^{3x} + 3e^{-4y} = C \) --- (1)
Applying the condition \( y = 0 \) when \( x = 0 \):
\( 4e^0 + 3e^0 = C \implies 4(1) + 3(1) = C \implies C = 7 \)
Substituting \( C = 7 \) into (1) gives the particular solution:
\( 4e^{3x} + 3e^{-4y} = 7 \)

Question. Find the particular solution of the differential equation \( x^2 \, dy = (2xy + y^2) \, dx \), given that \( y = 1 \), when \( x = 1 \).
Answer: Rewriting the equation:
\( \frac{dy}{dx} = \frac{2xy + y^2}{x^2} \) --- (1)
Let \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting into (1):
\( v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{x^2} = \frac{2vx^2 + v^2x^2}{x^2} = 2v + v^2 \)
\( \Rightarrow x\frac{dv}{dx} = 2v + v^2 - v = v^2 + v \)
Separating variables:
\( \frac{dv}{v(v+1)} = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{dv}{v(v+1)} = \int \frac{dx}{x} \)
Using partial fractions, \( \frac{1}{v(v+1)} = \frac{1}{v} - \frac{1}{v+1} \):
\( \int \left(\frac{1}{v} - \frac{1}{v+1}\right) \, dv = \int \frac{1}{x} \, dx \)
\( \Rightarrow \log|v| - \log|v+1| = \log|x| + \log C \)
\( \Rightarrow \log\left|\frac{v}{v+1}\right| = \log|Cx| \)
\( \Rightarrow \frac{v}{v+1} = Cx \)
Substituting \( v = \frac{y}{x} \):
\( \frac{\frac{y}{x}}{\frac{y}{x} + 1} = Cx \)
\( \Rightarrow \frac{y}{y+x} = Cx \)
\( \Rightarrow y = Cx(y+x) \) --- (2)
Applying the initial condition \( y = 1 \) when \( x = 1 \):
\( 1 = C(1)(1+1) \implies 1 = 2C \implies C = \frac{1}{2} \)
Substituting \( C = \frac{1}{2} \) into (2) gives the particular solution:
\( y = \frac{x(y+x)}{2} \)
\( \Rightarrow 2y = xy + x^2 \)

Question. Find the particular solution of the differential equation \( (1+x^2)\frac{dy}{dx} = \left(e^{m \tan^{-1} x} - y\right) \), given that \( y = 1 \) when \( x = 0 \).
Answer: Rearranging into standard linear form \( \frac{dy}{dx} + Py = Q \):
\( (1+x^2)\frac{dy}{dx} + y = e^{m\tan^{-1} x} \)
\( \Rightarrow \frac{dy}{dx} + \frac{1}{1+x^2} y = \frac{e^{m\tan^{-1} x}}{1+x^2} \)
Here, \( P = \frac{1}{1+x^2} \) and \( Q = \frac{e^{m\tan^{-1} x}}{1+x^2} \).
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int \frac{1}{1+x^2} \, dx} = e^{\tan^{-1} x} \)
The general solution is:
\( y \cdot e^{\tan^{-1} x} = \int \frac{e^{m\tan^{-1} x}}{1+x^2} \cdot e^{\tan^{-1} x} \, dx + C \)
\( \Rightarrow y e^{\tan^{-1} x} = \int \frac{e^{(m+1)\tan^{-1} x}}{1+x^2} \, dx + C \)
Let \( \tan^{-1} x = \theta \implies \frac{1}{1+x^2} \, dx = d\theta \):
\( \int e^{(m+1)\theta} \, d\theta = \frac{e^{(m+1)\theta}}{m+1} = \frac{e^{(m+1)\tan^{-1} x}}{m+1} \)
So the solution is:
\( y e^{\tan^{-1} x} = \frac{e^{(m+1)\tan^{-1} x}}{m+1} + C \)
\( \Rightarrow y = \frac{e^{m\tan^{-1} x}}{m+1} + C e^{-\tan^{-1} x} \) --- (1)
Applying the given condition \( y = 1 \) when \( x = 0 \):
\( 1 = \frac{e^0}{m+1} + C e^0 \)
\( \Rightarrow 1 = \frac{1}{m+1} + C \implies C = 1 - \frac{1}{m+1} = \frac{m}{m+1} \)
Substituting the value of \( C \) back into (1) gives the particular solution:
\( y = \frac{e^{m\tan^{-1} x}}{m+1} + \left(\frac{m}{m+1}\right) e^{-\tan^{-1} x} \)
\( \Rightarrow y = \frac{e^{m\tan^{-1} x} + m e^{-\tan^{-1} x}}{m+1} \)