CBSE Class 12 Mathematics Differential Equations Assignment Set 06

Long Answer Questions-I 

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: Given differential equation is \( \log \left( \frac{dy}{dx} \right) = 3x + 4y \)
\( \implies \) \( \frac{dy}{dx} = e^{3x + 4y} \)
\( \implies \) \( \frac{dy}{dx} = e^{3x} \cdot e^{4y} \)
\( \implies \) \( \frac{dy}{e^{4y}} = e^{3x} \cdot dx \)
\( \implies \) \( e^{-4y} dy = e^{3x} dx \)
Integrating both sides, we get
\( \int e^{-4y} dy = \int e^{3x} dx \)
\( \implies \) \( \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \)
\( \implies \) \( -3e^{-4y} = 4e^{3x} + 12C_1 \)
\( \implies \) \( 4e^{3x} + 3e^{-4y} = -12C_1 \)
\( \implies \) \( 4e^{3x} + 3e^{-4y} = C \dots (i) \)
It is general solution.
Now for particular solution we put \( x = 0 \) and \( y = 0 \) in (i), we get
\( 4 + 3 = C \)
\( \implies \) \( C = 7 \)
Putting \( C = 7 \) in (i), we get
\( 4e^{3x} + 3e^{-4y} = 7 \)
It is required particular solution.

Question. Solve the following differential equation:
\( 2x^2 \frac{dy}{dx} - 2xy + y^2 = 0 \) 
Answer: Given \( 2x^2 \frac{dy}{dx} - 2xy + y^2 = 0 \)
\( \implies \) \( 2x^2 \frac{dy}{dx} = 2xy - y^2 \)
\( \implies \) \( \frac{dy}{dx} = \frac{2xy - y^2}{2x^2} \dots (i) \)
It is homogeneous differential equation.
Let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Equation (i) becomes
\( v + x \frac{dv}{dx} = \frac{2x \cdot vx - v^2 x^2}{2x^2} \)
\( \implies \) \( v + x \frac{dv}{dx} = \frac{2x^2 \left( v - \frac{v^2}{2} \right)}{2x^2} \)
\( \implies \) \( x \frac{dv}{dx} = v - \frac{v^2}{2} - v \)
\( \implies \) \( x \frac{dv}{dx} = -\frac{v^2}{2} \)
\( \implies \) \( \frac{dx}{x} = -\frac{2dv}{v^2} \)
Integrating both sides, we get
\( \implies \) \( \int \frac{dx}{x} = -2 \int \frac{dv}{v^2} \)
\( \implies \) \( \log |x| + C = -2 \frac{v^{-2+1}}{-2+1} \)
\( \implies \) \( \log |x| + C = \frac{2}{v} \)
Putting \( v = \frac{y}{x} \), we get
\( \log |x| + C = \frac{2x}{y} \)

Question. Find the particular solution of the differential equation:
\( x(x^2 - 1) \frac{dy}{dx} = 1 \); \( y = 0 \); when \( x = 2 \)
Answer: Given differential equation is,
\( x(x^2 - 1) \frac{dy}{dx} = 1 \)
\( \implies \) \( dy = \frac{dx}{x(x^2 - 1)} \)
\( \implies \) \( dy = \frac{dx}{x(x - 1)(x + 1)} \)
Integrating both sides, we get,
\( \int dy = \int \frac{dx}{x(x - 1)(x + 1)} \)
\( \implies \) \( y = \int \frac{dx}{x(x - 1)(x + 1)} \dots (i) \)
Let \( \frac{1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \)
\( \implies \) \( \frac{1}{x(x - 1)(x + 1)} = \frac{A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)}{x(x - 1)(x + 1)} \)
\( \implies \) \( 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1) \)
Putting \( x = 1 \), we get \( 1 = 0 + B \cdot 1 \cdot 2 + 0 \)
\( \implies \) \( B = \frac{1}{2} \)
Putting \( x = -1 \), we get \( 1 = 0 + 0 + C \cdot (-1) \cdot (-2) \)
\( \implies \) \( C = \frac{1}{2} \)
Putting \( x = 0 \), we get \( 1 = A \cdot (-1) \cdot 1 \)
\( \implies \) \( A = -1 \)
Hence, \( \frac{1}{x(x - 1)(x + 1)} = \frac{-1}{x} + \frac{1}{2(x - 1)} + \frac{1}{2(x + 1)} \)
From (i) \( y = \int \left( -\frac{1}{x} + \frac{1}{2(x - 1)} + \frac{1}{2(x + 1)} \right) dx \)
\( \implies \) \( y = -\int \frac{dx}{x} + \frac{1}{2} \int \frac{dx}{x - 1} + \frac{1}{2} \int \frac{dx}{x + 1} \)
\( \implies \) \( y = -\log x + \frac{1}{2} \log |x - 1| + \frac{1}{2} \log |x + 1| + \log C_1 \)
\( \implies \) \( 2y = 2 \log \frac{1}{x} + \log |x^2 - 1| + 2 \log C_1 \)
\( \implies \) \( 2y = \log \left| \frac{x^2 - 1}{x^2} \right| + \log C_1^2 \dots (ii) \)
When \( x = 2 \), \( y = 0 \)
\( \implies \) \( 0 = \log \left| \frac{4 - 1}{4} \right| + \log C_1^2 \)
\( \implies \) \( \log C_1^2 = -\log \frac{3}{4} \)
Putting \( \log C_1^2 = -\log \frac{3}{4} \) in (ii), we get
\( 2y = \log \left| \frac{x^2 - 1}{x^2} \right| - \log \frac{3}{4} \)
\( \implies \) \( y = \frac{1}{2} \log \left| \frac{x^2 - 1}{x^2} \right| - \frac{1}{2} \log \frac{3}{4} \)

Question. Solve the differential equation \( (1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x} \) 
Answer: Given differential equation is
\( (1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x} \)
\( \implies \) \( \frac{dy}{dx} + \frac{1}{1 + x^2} y = \frac{e^{\tan^{-1} x}}{1 + x^2} \dots (i) \)
Equation (i) is of the form
\( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{1 + x^2} \), \( Q = \frac{e^{\tan^{-1} x}}{1 + x^2} \)
\( \therefore IF = e^{\int P dx} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} \)
Therefore, general solution of required differential equation is
\( y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \frac{e^{\tan^{-1} x}}{1 + x^2} dx + C \)
\( \implies \) \( y \cdot e^{\tan^{-1} x} = \int \frac{e^{2\tan^{-1} x}}{1 + x^2} dx + C \dots (ii) \)
Let \( \tan^{-1} x = z \)
\( \implies \) \( \frac{1}{1 + x^2} dx = dz \)
(ii) becomes
\( y \cdot e^{\tan^{-1} x} = \int e^{2z} dz + C \)
\( \implies \) \( y \cdot e^{\tan^{-1} x} = \frac{e^{2z}}{2} + C \)
\( \implies \) \( y \cdot e^{\tan^{-1} x} = \frac{e^{2\tan^{-1} x}}{2} + C \) [Putting \( z = \tan^{-1} x \)]
\( \implies \) \( y = \frac{e^{\tan^{-1} x}}{2} + C \cdot e^{-\tan^{-1} x} \) [Dividing both sides by \( e^{\tan^{-1} x} \)]
It is the required solution.

Question. 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: We have, \( e^x \sqrt{1 - y^2} dx + \frac{y}{x} dy = 0 \)
\( \implies \) \( e^x \sqrt{1 - y^2} dx = -\frac{y}{x} dy \)
\( \implies \) \( x e^x dx = -\frac{y}{\sqrt{1 - y^2}} dy \)
\( \implies \) \( \int x e^x dx = -\int \frac{y}{\sqrt{1 - y^2}} dy \)
\( \implies \) \( x e^x - \int e^x dx = \frac{1}{2} \int \frac{dt}{\sqrt{t}} \), where \( t = 1 - y^2 \) (Using ILATE on LHS)
\( \implies \) \( x e^x - e^x = \frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) + C \)
\( \implies \) \( x e^x - e^x = \sqrt{t} + C \)
\( \implies \) \( x e^x - e^x = \sqrt{1 - y^2} + C \), is the required solution.
Putting \( y = 1 \) and \( x = 0 \), we get
\( 0 e^0 - e^0 = \sqrt{1 - 1^2} + C \)
\( \implies \) \( C = -1 \)
Therefore, required particular solution is \( x e^x - e^x = \sqrt{1 - y^2} - 1 \).

Question. Solve the differential equation:
\( (\tan^{-1} y - x) dy = (1 + y^2) dx \) 
OR
Find the particular solution of the differential equation \( (\tan^{-1} y - x) dy = (1 + y^2) dx \), given that \( x = 1 \) when \( y = 0 \).
Answer: The given differential equation can be written as
\( \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2} \dots (i) \)
Now, (i) is of the form \( \frac{dx}{dy} + Px = Q \), where \( P = \frac{1}{1 + y^2} \) and \( Q = \frac{\tan^{-1} y}{1 + y^2} \)
Therefore, IF = \( e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y} \)
Thus, the solution of the given differential equation is
\( x e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1 + y^2} \right) e^{\tan^{-1} y} dy + C \dots (ii) \)
Let \( I = \int \left( \frac{\tan^{-1} y}{1 + y^2} \right) e^{\tan^{-1} y} dy \)
Substituting \( \tan^{-1} y = t \) so that \( \left( \frac{1}{1 + y^2} \right) dy = dt \), we get
\( I = \int t e^t dt = t e^t - \int 1 \cdot e^t dt = t e^t - e^t \equiv e^t(t - 1) \)
or \( I = e^{\tan^{-1} y}(\tan^{-1} y - 1) \)
Substituting the value of \( I \) in equation (ii), we get
\( x \cdot e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C \)
or \( x = (\tan^{-1} y - 1) + C e^{-\tan^{-1} y} \) is the required solution.
OR
For general solution take the above solution.
General solution is \( x = (\tan^{-1} y - 1) + C e^{-\tan^{-1} y} \)
For particular solution putting \( x = 1, y = 0 \), we get
\( 1 = (\tan^{-1} 0 - 1) + C e^{-\tan^{-1} 0} \)
\( 1 = -1 + C \)
\( \implies \) \( C = 2 \)
Therefore required particular solution is
\( x = (\tan^{-1} y - 1) + 2 e^{-\tan^{-1} y} \)

Question. Solve the following differential equation:
\( \left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1, x \neq 0 \) 
Answer: Given \( \left( \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right) \frac{dx}{dy} = 1, x \neq 0 \)
\( \implies \) \( \frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \)
\( \implies \) \( \frac{dy}{dx} + \frac{1}{\sqrt{x}} \cdot y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \)
It is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{\sqrt{x}}, Q = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \)
\( \therefore IF = e^{\int P dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{\int x^{-1/2} dx} = e^{\frac{x^{-1/2 + 1}}{-1/2 + 1}} = e^{2\sqrt{x}} \)
Therefore general solution is
\( y \cdot e^{2\sqrt{x}} = \int Q \times \text{IF} dx + C \)
\( \implies \) \( y \cdot e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} dx + C \)
\( \implies \) \( y \cdot e^{2\sqrt{x}} = \int \frac{dx}{\sqrt{x}} + C \)
\( \implies \) \( y \cdot e^{2\sqrt{x}} = \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C \
\( \implies \) \( y \cdot e^{2\sqrt{x}} = 2\sqrt{x} + C \)

Question. Solve the differential equation \( (x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1} \), where \( x \in (-\infty, -1) \cup (1, \infty) \).
Answer: The given differential equation is \( (x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1} \)
\( \implies \) \( \frac{dy}{dx} + \frac{2x}{x^2 - 1} y = \frac{2}{(x^2 - 1)^2} \dots (i) \)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2x}{x^2 - 1} \) and \( Q = \frac{2}{(x^2 - 1)^2} \)
\( \therefore IF = e^{\int P dx} = e^{\int 2x / (x^2 - 1) dx} = e^{\log(x^2 - 1)} = x^2 - 1 \)
Multiplying both sides of (i) by IF = \( x^2 - 1 \), we get \( (x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1} \)
Integrating both sides, we get
\( y(x^2 - 1) = \int \frac{2}{x^2 - 1} dx + C \) [Using \( y(\text{IF}) = \int Q \cdot (\text{IF}) dx + C \)]
\( \implies \) \( y(x^2 - 1) = 2 \cdot \frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right| + C \)
\( \implies \) \( y(x^2 - 1) = \log \left| \frac{x - 1}{x + 1} \right| + C \)
This is the required solution.

Question. Find the particular solution of the differential equation \( \frac{dy}{dx} = 1 + x + y + xy \) given that \( y = 0 \) when \( x = 1 \). 
Answer: Given differential equation is \( \frac{dy}{dx} = 1 + x + y + xy \)
\( \implies \) \( \frac{dy}{dx} = (1 + x) + y(1 + x) \)
\( \implies \) \( \frac{dy}{dx} = (1 + x)(1 + y) \)
\( \implies \) \( \frac{dy}{1 + y} = (1 + x)dx \)
Integrating both sides, we get \( \log |1 + y| = \int (1 + x) dx \)
\( \implies \) \( \log |1 + y| = x + \frac{x^2}{2} + C \) is the general solution.
Putting \( x = 1, y = 0 \), we get
\( \log 1 = 1 + \frac{1}{2} + C \)
\( \implies \) \( 0 = \frac{3}{2} + C \)
\( \implies \) \( C = -\frac{3}{2} \)
Hence, particular solution is \( \log |1 + y| = x + \frac{x^2}{2} - \frac{3}{2} \).

Question. Solve the differential equation \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \). 
Answer: Given differential equation is \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \)
\( \implies \) \( \frac{dy}{dx} + \left( \frac{1}{x \cdot \log x} \right) \cdot y = \frac{2}{x^2} \) (Divide each term by \( x \log x \))
It is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{x \cdot \log x} \) and \( Q = \frac{2}{x^2} \).
\( \therefore IF = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} \)
Put \( \log x = z \)
\( \implies \) \( \frac{dx}{x} = dz \)
\( = e^{\int \frac{1}{z} dz} = e^{\log z} = z = \log x \)
\( \therefore \) General solution is
\( y \cdot \log x = \int \log x \cdot \frac{2}{x^2} dx + C \)
\( \implies \) \( y \log x = 2 \int \frac{\log x}{x^2} dx + C \)
Let \( \log x = z \)
\( \implies \) \( \frac{1}{x} dx = dz \) also \( \log x = z \)
\( \implies \) \( x = e^z \)
\( \therefore y \log x = 2 \int \frac{z}{e^z} dz + C \)
\( \implies \) \( y \log x = 2 \int z \cdot e^{-z} dz + C \)
\( \implies \) \( y \log x = 2 \left[ z \cdot \frac{e^{-z}}{-1} - \int 1 \cdot \frac{e^{-z}}{-1} dz \right] + C \)
\( \implies \) \( y \log x = 2[-z e^{-z} + \int e^{-z} dz] + C \)
\( \implies \) \( y \log x = -2z e^{-z} - 2e^{-z} + C \)
\( \implies \) \( y \log x = -2\log x e^{-\log x} - 2e^{-\log x} + C \)
\( \implies \) \( y \log x = -2\log x \cdot \frac{1}{x} - \frac{2}{x} + C \)
\( \left[ \because e^{-\log x} = e^{\log \frac{1}{x}} = \frac{1}{x} \right] \)
\( \implies \) \( y \log x = -\frac{2}{x}(1 + \log x) + C \)

Question. Show that the differential equation \( x \frac{dy}{dx} \sin\left(\frac{y}{x}\right) + x - y \sin\left(\frac{y}{x}\right) = 0 \) is homogeneous. Find the particular solution of this differential equation, given that \( x = 1 \) when \( y = \frac{\pi}{2} \). 
Answer: Given differential equation is \( x \frac{dy}{dx} \sin\left(\frac{y}{x}\right) + x - y \sin\left(\frac{y}{x}\right) = 0 \)
Dividing both sides by \( x \sin\left(\frac{y}{x}\right) \), we get
\( \frac{dy}{dx} + \text{cosec}\left(\frac{y}{x}\right) - \frac{y}{x} = 0 \)
\( \implies \) \( \frac{dy}{dx} = \frac{y}{x} - \text{cosec}\left(\frac{y}{x}\right) \dots (i) \)
Let \( F(x, y) = \frac{y}{x} - \text{cosec}\left(\frac{y}{x}\right) \)
\( \therefore F(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \text{cosec}\left(\frac{\lambda y}{\lambda x}\right) = \lambda^0 \left[ \frac{y}{x} - \text{cosec}\left(\frac{y}{x}\right) \right] = \lambda^0 F(x, y) \)
Hence, differential equation (i) is homogeneous.
Let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \cdot \frac{dv}{dx} \)
Now, equation (i) becomes
\( v + x \cdot \frac{dv}{dx} = \frac{vx}{x} - \text{cosec}\left(\frac{vx}{x}\right) \)
\( v + x \cdot \frac{dv}{dx} = v - \text{cosec} \, v \)
\( \implies \) \( x \cdot \frac{dv}{dx} = -\text{cosec} \, v \)
\( \implies \) \( -\sin v \, dv = \frac{dx}{x} \)
\( \implies \) \( -\int \sin v \, dv = \int \frac{dx}{x} \)
\( \implies \) \( \cos v = \log |x| + C \)
\( \implies \) \( \cos\left(\frac{y}{x}\right) = \log |x| + C \dots (ii) \)
Putting \( y = \frac{\pi}{2}, x = 1 \) in (ii), we get
\( \therefore \cos\left(\frac{\pi}{2}\right) = \log 1 + C \)
\( \implies \) \( 0 = 0 + C \)
\( \implies \) \( C = 0 \)
Hence, particular solution is
\( \cos\left(\frac{y}{x}\right) = \log |x| + 0 \)
i.e., \( \cos\left(\frac{y}{x}\right) = \log |x| \)

Question. Solve the following differential equation:
\( \cos^2 x \frac{dy}{dx} + y = \tan x \) 
Answer: Given differential equation is,
\( \cos^2 x \cdot \frac{dy}{dx} + y = \tan x \)
\( \implies \) \( \frac{dy}{dx} + y \sec^2 x = \tan x \cdot \sec^2 x \)
Given differential equation is of the type \( \frac{dy}{dx} + Py = Q \), where \( P = \sec^2 x \) and \( Q = \tan x \cdot \sec^2 x \).
IF = \( e^{\int P dx} = e^{\int \sec^2 x dx} = e^{\tan x} \)
\( \therefore \) Solution is given by \( e^{\tan x} y = \int \tan x \cdot \sec^2 x \cdot e^{\tan x} dx \)
Let \( I = \int \tan x \cdot \sec^2 x \cdot e^{\tan x} dx \)
Put \( \tan x = t, \sec^2 x \, dx = dt \), we get
\( I = \int t \, e^t dt \)
\( \therefore = t e^t - \int e^t dt = t e^t - e^t + C = \tan x \, e^{\tan x} - e^{\tan x} + C \) [Integrating by parts]
Hence, \( e^{\tan x} y = e^{\tan x}(\tan x - 1) + C \)
\( \implies \) \( y = \tan x - 1 + C \, e^{-\tan x} \)

Question. Solve the differential equation:
\( \sqrt{1 + x^2 + y^2 + x^2 y^2} + xy \frac{dy}{dx} = 0 \)
Answer: Given \( \sqrt{1 + x^2 + y^2 + x^2 y^2} + xy \frac{dy}{dx} = 0 \)
By simplifying the equation, we get
\( xy \frac{dy}{dx} = -\sqrt{1 + x^2 + y^2 + x^2 y^2} = -\sqrt{1 + x^2 + y^2(1 + x^2)} \)
\( \implies \) \( xy \frac{dy}{dx} = -\sqrt{(1 + x^2)(1 + y^2)} = -\sqrt{1 + x^2} \sqrt{1 + y^2} \)
\( \implies \) \( \frac{y}{\sqrt{1 + y^2}} dy = -\frac{\sqrt{1 + x^2}}{x} dx \)
Integrating both sides, we get
\( \int \frac{y}{\sqrt{(1 + y^2)}} dy = -\int \frac{\sqrt{(1 + x^2)}}{x} dx \dots (i) \)
Let \( 1 + y^2 = t \)
\( \implies \) \( 2y dy = dt \) and \( 1 + x^2 = m^2 \)
\( \implies \) \( 2x dx = 2m dm \)
\( \implies \) \( x dx = m dm \)
\( \therefore (i) \implies \frac{1}{2} \int \frac{1}{\sqrt{t}} dt = -\int \frac{m}{m^2 - 1} \cdot m dm \)
\( \implies \) \( \frac{1}{2} \frac{t^{1/2}}{1/2} + \int \frac{m^2}{m^2 - 1} dm = 0 \)
\( \implies \) \( \sqrt{t} + \int \frac{m^2 + 1 - 1}{m^2 - 1} dm = 0 \)
\( \implies \) \( \sqrt{t} + \int \left( 1 + \frac{1}{m^2 - 1} \right) dm = 0 \)
\( \implies \) \( \sqrt{t} + m + \frac{1}{2} \log \left| \frac{m - 1}{m + 1} \right| = 0 \)
Now, substituting these value of t and m, we get
\( \sqrt{1 + y^2} + \sqrt{1 + x^2} + \frac{1}{2} \log \left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right| + C = 0 \)

Question. Solve the differential equation:
\( (x^2 + 1) \frac{dy}{dx} + 2xy = \sqrt{x^2 + 4} \)
Answer: We have \( (x^2 + 1) \frac{dy}{dx} + 2xy = \sqrt{x^2 + 4} \)
Simplifying the above equation, we get
\( \frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{\sqrt{x^2 + 4}}{(x^2 + 1)} \)
This is of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2x}{x^2 + 1}, Q = \frac{\sqrt{x^2 + 4}}{(x^2 + 1)} \).
Now, IF = \( e^{\int P dx} = e^{\int \frac{2x}{x^2 + 1} dx} = e^{\log(x^2 + 1)} = (x^2 + 1) \)
Its solution is given by
\( (x^2 + 1)y = \int (x^2 + 1) \cdot \frac{\sqrt{x^2 + 4}}{(x^2 + 1)} dx = \int \sqrt{x^2 + 4} dx \)
\( \implies \) \( (x^2 + 1)y = \frac{x}{2} \sqrt{x^2 + 4} + \frac{4}{2} \log |x + \sqrt{x^2 + 4}| + C \)

Question. \( (x^2 + y^2) dy = xy dx \). If \( y(1) = 1 \) and \( y(x_0) = e \), then find the value of \( x_0 \). 
Answer: Given differential equation is \( (x^2 + y^2) dy = xy dx \)
It is also written as
\( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \dots (i) \)
Now, to solve let \( y = vx \).
Differentiating \( y = vx \) with respect to x, we get
\( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \)
Putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in (i), we get
\( v + x \frac{dv}{dx} = \frac{x \cdot vx}{x^2 + (vx)^2} \)
\( v + x \frac{dv}{dx} = \frac{vx^2}{x^2 + v^2 x^2} \)
\( \implies \) \( v + x \frac{dv}{dx} = \frac{vx^2}{x^2(1 + v^2)} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{v}{(1 + v^2)} - v \)
\( \implies \) \( x \frac{dv}{dx} = \frac{v - v - v^3}{(1 + v^2)} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{-v^3}{(1 + v^2)} \)
\( \implies \) \( \frac{(1 + v^2) dv}{v^3} = -\frac{dx}{x} \)
Integrating both sides, we get
\( \int \frac{(1 + v^2) dv}{v^3} = -\int \frac{dx}{x} \)
\( \implies \) \( \int \frac{dv}{v^3} + \int \frac{dv}{v} = -\log |x| + C \)
\( \implies \) \( -\frac{1}{2v^2} + \log |v| = -\log |x| + C \)
\( \implies \) \( -\frac{x^2}{2y^2} + \log \left| \frac{y}{x} \right| = -\log |x| + C \)
\( \implies \) \( -\frac{x^2}{2y^2} + \log |y| - \log |x| = -\log |x| + C \)
\( \implies \) \( -\frac{x^2}{2y^2} + \log |y| = C \dots (ii) \)
Given, \( x = 1, y = 1 \)
\( \implies \) \( -\frac{1}{2 \times 1} + \log |1| = C \)
\( \implies \) \( -\frac{1}{2} = C \) \( [\because \log 1 = 0] \)
Now (ii) becomes
\( -\frac{x^2}{2y^2} + \log |y| = -\frac{1}{2} \)
\( \implies \) \( \log |y| = \frac{x^2}{2y^2} - \frac{1}{2} \)
\( \implies \) \( \log |y| = \frac{x^2 - y^2}{2y^2} \dots (iii) \)
Putting \( x = x_0 \) and \( y = e \) in (iii), we get
\( \log |e| = \frac{x_0^2 - e^2}{2e^2} \)
\( \implies \) \( 1 = \frac{x_0^2 - e^2}{2e^2} \)
\( \implies \) \( x_0^2 - e^2 = 2e^2 \)
\( \implies \) \( x_0^2 = 3e^2 \)
\( \implies \) \( x_0 = \sqrt{3}e \)

Question. Find the particular solution of the differential equation.
\( \frac{dy}{dx} + y \tan x = 3x^2 + x^3 \tan x, x \neq \frac{\pi}{2} \), given that \( y = 0 \) when \( x = \frac{\pi}{3} \). 
Answer: Given, \( \frac{dy}{dx} + y \tan x = 3x^2 + x^3 \tan x \)
\( \implies \) \( \frac{dy}{dx} + \tan x \cdot y = 3x^2 + x^3 \tan x \)
This is of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \tan x, Q = 3x^2 + x^3 \tan x \).
\( \therefore IF = e^{\int \tan x dx} = e^{\log \sec x} = \sec x \)
Therefore, general solution is given by
\( y \cdot \sec x = \int (3x^2 + x^3 \tan x) \cdot \sec x \, dx + C \)
\( \implies \) \( y \cdot \sec x = \int 3x^2 \sec x \, dx + \int x^3 \tan x \cdot \sec x \, dx + C \)
\( \implies \) \( y \sec x = \int 3x^2 \sec x \, dx + x^3 \cdot \sec x - \int 3x^2 \cdot \sec x \, dx + C \)
\( \implies \) \( y \sec x = x^3 \sec x + C \)
\( \implies \) \( y = x^3 + C \cos x \)
Now \( x = \frac{\pi}{3}, y = 0 \)
\( \therefore 0 = \left(\frac{\pi}{3}\right)^3 + C \cdot \cos\left(\frac{\pi}{3}\right) \)
\( \implies \) \( 0 = \frac{\pi^3}{27} + \frac{C}{2} \)
\( \implies \) \( C = -\frac{2\pi^3}{27} \)
Hence, required particular solution is \( y = x^3 - \frac{2\pi^3}{27} \cos x \).

Question. Show that the differential equation \( (x - y) \frac{dy}{dx} = x + 2y \) is homogeneous and solve it.
OR
Show that \( (x - y) dy = (x + 2y) dx \) is a homogenous differential equation. Also, find the general solution of the given differential equation.
Answer: Given, \( (x - y) \frac{dy}{dx} = x + 2y \)
By simplifying the above equation, we get
\( \frac{dy}{dx} = \frac{x + 2y}{x - y} \dots (i) \)
Let \( F(x, y) = \frac{x + 2y}{x - y} \)
then \( F(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + 2y)}{\lambda(x - y)} = \lambda^0 F(x, y) \)
\( F(x, y) \) is homogeneous function and hence given differential equation is homogeneous.
Now, let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Substituting these values in equation (i), we get
\( v + x \frac{dv}{dx} = \frac{x + 2vx}{x - vx} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v + v^2}{1 - v} = \frac{1 + v + v^2}{1 - v} \)
\( \implies \) \( \frac{1 - v}{1 + v + v^2} dv = \frac{dx}{x} \)
By integrating both sides, we get
\( \int \frac{1 - v}{1 + v + v^2} dv = \int \frac{dx}{x} \dots (ii) \)
\( \text{LHS} = \int \frac{1 - v}{v^2 + v + 1} dv \)
Let \( 1 - v = A(2v + 1) + B = 2Av + (A + B) \)
Comparing coefficients of both sides, we get
\( 2A = -1, A + B = 1 \)
or \( A = -\frac{1}{2}, B = \frac{3}{2} \)
\( \therefore \int \frac{1 - v}{v^2 + v + 1} dv = \int \frac{-\frac{1}{2}(2v + 1) + \frac{3}{2}}{v^2 + v + 1} dv \)
\( = -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} dv + \frac{3}{2} \int \frac{dv}{v^2 + v + 1} \)
\( = -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} dv + \frac{3}{2} \int \frac{dv}{\left(v + \frac{1}{2}\right)^2 + \frac{3}{4}} \)
\( = -\frac{1}{2} \log|v^2 + v + 1| + \frac{3}{2} \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \)
Now, substituting it in equation (ii), we get
\( -\frac{1}{2} \log|v^2 + v + 1| + \sqrt{3} \tan^{-1} \left( \frac{2v + 1}{\sqrt{3}} \right) = \log x + C \)
\( \implies \) \( -\frac{1}{2} \log\left| \frac{y^2}{x^2} + \frac{y}{x} + 1 \right| + \sqrt{3} \tan^{-1} \left( \frac{\frac{2y}{x} + 1}{\sqrt{3}} \right) = \log x + C \)
\( \implies \) \( -\frac{1}{2} \log\left| \frac{y^2 + xy + x^2}{x^2} \right| + \sqrt{3} \tan^{-1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log x + C \)
\( \implies \) \( -\frac{1}{2} \log |x^2 + xy + y^2| + \frac{1}{2} \log x^2 + \sqrt{3} \tan^{-1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log x + C \)
\( \implies \) \( -\frac{1}{2} \log |x^2 + xy + y^2| + \sqrt{3} \tan^{-1} \left( \frac{2y + x}{\sqrt{3}x} \right) = C \)

Question. Solve \( \frac{dy}{dx} = \cos(x + y) + \sin(x + y) \). 
Answer: Given, \( \frac{dy}{dx} = \cos(x + y) + \sin(x + y) \)
Put \( x + y = z \)
\( \implies \) \( 1 + \frac{dy}{dx} = \frac{dz}{dx} \)
On substituting these values in equation (i), we get
\( \left( \frac{dz}{dx} - 1 \right) = \cos z + \sin z \)
\( \implies \) \( \frac{dz}{dx} = (\cos z + \sin z + 1) \)
\( \implies \) \( \frac{dz}{\cos z + \sin z + 1} = dx \)
On integrating both sides, we get
\( \int \frac{dz}{\cos z + \sin z + 1} = \int 1 dx \)
\( \implies \) \( \int \frac{dz}{\frac{1 - \tan^2(z/2)}{1 + \tan^2(z/2)} + \frac{2\tan(z/2)}{1 + \tan^2(z/2)} + 1} = \int dx \)
\( \implies \) \( \int \frac{dz}{\frac{1 - \tan^2(z/2) + 2\tan(z/2) + 1 + \tan^2(z/2)}{(1 + \tan^2(z/2))}} = \int dx \)
\( \implies \) \( \int \frac{(1 + \tan^2(z/2)) dz}{2 + 2\tan(z/2)} = \int dx \)
\( \implies \) \( \int \frac{\sec^2(z/2) dz}{2(1 + \tan(z/2))} = \int dx \)
Put \( 1 + \tan(z/2) = t \)
\( \implies \) \( \left( \frac{1}{2} \sec^2(z/2) \right) dz = dt \)
\( \implies \) \( \int \frac{dt}{t} = \int dx \)
\( \implies \) \( \log|t| = x + C \)
\( \implies \) \( \log|1 + \tan(z/2)| = x + C \)
\( \implies \) \( \log\left| 1 + \tan\left( \frac{x + y}{2} \right) \right| = x + C \)

Question. Find the equation of the curve through the point (1, 0), if the slope of the tangent to the curve at any point (x, y) is \( \frac{y - 1}{x^2 + x} \). 
Answer: It is given that, slope of tangent to the curve at any point (x, y) is \( \frac{y - 1}{x^2 + x} \).
\( \therefore \left( \frac{dy}{dx} \right)_{(x, y)} = \frac{y - 1}{x^2 + x} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y - 1}{x^2 + x} \)
\( \implies \) \( \frac{dy}{y - 1} = \frac{dx}{x^2 + x} \)
On integrating both sides, we get \( \int \frac{dy}{y - 1} = \int \frac{dx}{x^2 + x} \)
\( \implies \) \( \int \frac{dy}{y - 1} = \int \frac{dx}{x(x + 1)} \)
\( \implies \) \( \int \frac{dy}{y - 1} = \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx \)
\( \implies \) \( \log(y - 1) = \log x - \log(x + 1) + \log C \)
\( \implies \) \( \log(y - 1) = \log\left( \frac{xC}{x + 1} \right) \)
Since, the given curve passes through point (1, 0).
\( \therefore 0 - 1 = \frac{1 \cdot C}{1 + 1} \)
\( \implies \) \( C = -2 \)
The particular solution is \( y - 1 = \frac{-2x}{x + 1} \)
\( \implies \) \( (y - 1)(x + 1) = -2x \)
\( \implies \) \( (y - 1)(x + 1) + 2x = 0 \).

Question. Find the particular solution of the differential equation:
\( (1 - y^2)(1 + \log x) dx + 2xy dy = 0 \) given that \( y = 0 \) when \( x = 1 \) 
Answer: We have \( (1 - y^2)(1 + \log x) dx + 2xy dy = 0 \)
\( \implies \) \( 2xy dy = -(1 - y^2)(1 + \log x) dx \)
\( \implies \) \( \frac{2y dy}{1 - y^2} = -\frac{(1 + \log x) dx}{x} \)
Integrating both sides, we get
\( \implies \) \( \int \frac{2y}{1 - y^2} dy = -\int \frac{(1 + \log x)}{x} dx \)
\( \implies \) \( -\log |1 - y^2| = -\int \frac{(1 + \log x)}{x} dx \)
\( \implies \) \( -\log |1 - y^2| = -\int z dz \) [Let \( 1 + \log x = z \)
\( \implies \) \( \frac{1}{x} dx = dz \)]
\( \implies \) \( \log |1 - y^2| = \frac{z^2}{2} + C \)
\( \implies \) \( \log |1 - y^2| = \frac{(1 + \log x)^2}{2} + C \)
Putting \( x = 1 \) and \( y = 0 \), we get
\( \implies \) \( \log 1 = \frac{(1 + \log 1)^2}{2} + C \)
\( \implies \) \( 0 = \frac{1}{2} + C \)
\( \implies \) \( C = -\frac{1}{2} \)
Hence, particular solution is \( \log |1 - y^2| = \frac{(1 + \log x)^2}{2} - \frac{1}{2} \).

Question. Find the general solution of the following differential equation:
\( (1 + y^2) + (x - e^{\tan^{-1} y}) \frac{dy}{dx} = 0 \) 
Answer: We have \( (1 + y^2) + (x - e^{\tan^{-1} y}) \frac{dy}{dx} = 0 \)
\( \implies \) \( (x - e^{\tan^{-1} y}) \frac{dy}{dx} = -(1 + y^2) \)
\( \implies \) \( \frac{dy}{dx} = -\frac{1 + y^2}{x - e^{\tan^{-1} y}} \)
\( \implies \) \( \frac{dx}{dy} = \frac{x - e^{\tan^{-1} y}}{-(1 + y^2)} \)
\( \implies \) \( \frac{dx}{dy} = -\frac{x}{1 + y^2} + \frac{e^{\tan^{-1} y}}{1 + y^2} \)
\( \implies \) \( \frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{e^{\tan^{-1} y}}{1 + y^2} \)
It is in the form \( \frac{dx}{dy} + Px = Q \), where \( P = \frac{1}{1 + y^2} \) and \( Q = \frac{e^{\tan^{-1} y}}{1 + y^2} \).
\( \therefore IF = e^{\int P dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y} \)
Therefore, general solution is \( x \cdot e^{\tan^{-1} y} = \int \frac{e^{\tan^{-1} y}}{1 + y^2} \cdot e^{\tan^{-1} y} dy + C \)
\( \implies \) \( x \cdot e^{\tan^{-1} y} = \int e^z \cdot e^z dz + C \)
[Let \( \tan^{-1} y = z \implies \frac{1}{1 + y^2} dy = dz \)]
\( \implies \) \( x \cdot e^{\tan^{-1} y} = \int e^{2z} dz + C \)
\( \implies \) \( x \cdot e^{\tan^{-1} y} = \frac{e^{2z}}{2} + C \)
\( \implies \) \( x \cdot e^{\tan^{-1} y} = \frac{e^{2\tan^{-1} y}}{2} + C \)
\( \implies \) \( x = \frac{1}{2} e^{\tan^{-1} y} + C \cdot e^{-\tan^{-1} y} \)

Question. Find the particular solution of differential equation: \( \frac{dy}{dx} = -\frac{x + y \cos x}{1 + \sin x} \) given that \( y = 1 \) when \( x = 0 \). 
Answer: We have
\( \frac{dy}{dx} = -\frac{x + y \cos x}{1 + \sin x} \)
\( \implies \) \( \frac{dy}{dx} = -\frac{x}{1 + \sin x} - \frac{y \cos x}{1 + \sin x} \)
\( \implies \) \( \frac{dy}{dx} + \frac{\cos x}{1 + \sin x} y = -\frac{x}{1 + \sin x} \)
It is in the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{\cos x}{1 + \sin x}, Q = -\frac{x}{1 + \sin x} \).
Now \( IF = e^{\int \frac{\cos x}{1 + \sin x} dx} = e^{\log|1 + \sin x|} = 1 + \sin x \)
Therefore, general solution is
\( y(1 + \sin x) = \int -\frac{x}{1 + \sin x}(1 + \sin x) dx + C = -\int x dx + C \)
\( \implies \) \( y(1 + \sin x) = -\frac{x^2}{2} + C \)
\( 1(1 + \sin 0) = 0 + C \)
\( \implies \) \( C = 1 \) [Given \( y = 1 \) and \( x = 0 \)]
Hence, particular solution is
\( y(1 + \sin x) = -\frac{x^2}{2} + 1 \)
\( \implies \) \( y = \frac{2 - x^2}{2(1 + \sin x)} \)

Question. Solve the following differential equation:
\( (\cot^{-1} y + x)dy = (1 + y^2)dx \) 
Answer: We have \( (\cot^{-1} y + x)dy = (1 + y^2)dx \)
This can be written as
\( \implies \) \( \frac{dx}{dy} = \frac{\cot^{-1} y + x}{1 + y^2} = \frac{\cot^{-1} y}{1 + y^2} + \frac{x}{1 + y^2} \)
\( \implies \) \( \frac{dx}{dy} - \frac{1}{1 + y^2} \cdot x = \frac{\cot^{-1} y}{1 + y^2} \)
It is of the form \( \frac{dx}{dy} + Px = Q \), where \( P = \frac{-1}{1 + y^2} \) and \( Q = \frac{\cot^{-1} y}{1 + y^2} \).
\( \therefore IF = e^{\int \frac{-1}{1 + y^2} dy} = e^{\cot^{-1} y} \)
Therefore, required solution of differential equation is
\( x \cdot e^{\cot^{-1} y} = \int \frac{\cot^{-1} y}{1 + y^2} \cdot e^{\cot^{-1} y} dy + C \)
\( \implies \) \( x \cdot e^{\cot^{-1} y} = I + C \dots (i) \)
Here, \( I = \int \frac{\cot^{-1} y}{1 + y^2} \cdot e^{\cot^{-1} y} dy \)
Let \( \cot^{-1} y = t \)
\( \implies \) \( -\frac{1}{1 + y^2} dy = dt \)
\( \implies \) \( \frac{1}{1 + y^2} dy = -dt \)
\( \implies \) \( I = -\int t \cdot e^t dt = -[t \cdot e^t - \int e^t dt] = -t \cdot e^t + e^t = e^t(1 - t) = e^{\cot^{-1} y}(1 - \cot^{-1} y) \)
Hence, required solution is \( x \cdot e^{\cot^{-1} y} = e^{\cot^{-1} y}(1 - \cot^{-1} y) + C \). [From equation (i)]
\( \implies \) \( x = (1 - \cot^{-1} y) + C e^{-\cot^{-1} y} \)

Question. Find the general solution of the following differential equation:
\( x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \) 
Answer: Given differential equation is \( x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \)
\( \frac{dy}{dx} = \frac{y \cos(y/x) + x}{x \cos(y/x)} \dots (i) \)
It is homogeneous differential equation.
Let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
(i) \( \implies v + x \frac{dv}{dx} = \frac{vx \cos v + x}{x \cdot \cos v} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{v \cos v + 1}{\cos v} - v \)
\( \implies \) \( x \frac{dv}{dx} = \frac{v \cos v + 1 - v \cos v}{\cos v} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{1}{\cos v} \)
\( \implies \) \( \cos v dv = \frac{dx}{x} \)
Integrating both sides
\( \implies \) \( \sin v = \log|x| + C \)
\( \implies \) \( \sin \frac{y}{x} = \log|x| + C \) is the required solution.

Long Answer Questions-II 

Question. Solve the following differential equation:
\( 3e^x \tan y dx + (2 - e^x) \sec^2 y dy = 0 \), given that when \( x = 0, y = \frac{\pi}{4} \) 
Answer: Given, \( 3e^x \tan y dx + (2 - e^x) \sec^2 y dy = 0 \)
\( \implies \) \( (2 - e^x) \sec^2 y dy = -3e^x \tan y dx \)
\( \implies \) \( \frac{\sec^2 y}{\tan y} dy = \frac{-3e^x}{2 - e^x} dx \)
\( \implies \) \( \int \frac{\sec^2 y dy}{\tan y} = 3 \int \frac{-e^x dx}{2 - e^x} \)
\( \implies \) \( \log|\tan y| = 3\log|2 - e^x| + \log C \)
\( \implies \) \( \log|\tan y| = \log|C \cdot (2 - e^x)^3| \)
\( \implies \) \( \tan y = C(2 - e^x)^3 \)
Putting \( x = 0, y = \frac{\pi}{4} \), we get
\( \implies \) \( \tan \frac{\pi}{4} = C(2 - e^0)^3 \)
\( \implies \) \( 1 = C(2 - 1)^3 \)
\( \implies \) \( 1 = C \)
Therefore, particular solution is \( \tan y = (2 - e^x)^3 \).

Question. Solve: \( x dy - y dx = \sqrt{x^2 + y^2} dx \) 
Answer: The given differential equation can be written as
\( \frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x}, x \neq 0 \)
Clearly, it is a homogeneous differential equation.
Putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in it, we get
\( v + x \frac{dv}{dx} = \frac{\sqrt{x^2 + v^2 x^2} + vx}{x} \)
\( \implies \) \( v + x \frac{dv}{dx} = \sqrt{1 + v^2} + v \)
\( \implies \) \( x \frac{dv}{dx} = \sqrt{1 + v^2} \)
\( \implies \) \( \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \)
Integrating both sides, we get
\( \int \frac{1}{\sqrt{1 + v^2}} dv = \int \frac{1}{x} dx \)
\( \implies \) \( \log|v + \sqrt{1 + v^2}| = \log|x| + \log C \)
\( \implies \) \( |v + \sqrt{1 + v^2}| = |Cx| \)
\( \implies \) \( \left| \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} \right| = |Cx| \) \( [\because v = y/x] \)
\( \implies \) \( \{y + \sqrt{x^2 + y^2}\}^2 = C^2 x^4 \) [Squaring both sides]
Hence, \( \{y + \sqrt{x^2 + y^2}\}^2 = C^2 x^4 \) gives the required solution.

Question. Show that the differential equation \( (x e^{y/x} + y) dx = x dy \) is homogeneous. Find the particular solution of this differential equation, given that \( x = 1 \) when \( y = 1 \). 
Answer: Given differential equation is \( \left( x \cdot e^{y/x} + y \right) dx = x dy \)
\( \implies \) \( \frac{dy}{dx} = \frac{x \cdot e^{y/x} + y}{x} \dots (i) \)
Let \( F(x, y) = \frac{x \cdot e^{y/x} + y}{x} \)
\( \implies \) \( F(\lambda x, \lambda y) = \frac{\lambda x \cdot e^{\lambda y / \lambda x} + \lambda y}{\lambda x} = \lambda^0 \frac{x \cdot e^{y/x} + y}{x} = \lambda^0 F(x, y) \)
Hence, given differential equation (i) is homogeneous.
Let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \cdot \frac{dv}{dx} \)
Now, given differential equation (i) would become
\( v + x \cdot \frac{dv}{dx} = \frac{x \cdot e^v + vx}{x} \)
\( \implies \) \( v + x \cdot \frac{dv}{dx} = e^v + v \)
\( \implies \) \( x \cdot \frac{dv}{dx} = e^v \)
\( \implies \) \( \frac{dv}{e^v} = \frac{dx}{x} \)
\( \implies \) \( \int e^{-v} dv = \int \frac{dx}{x} \)
\( \implies \) \( \frac{e^{-v}}{-1} = \log x + C \)
\( \implies \) \( -e^{-y/x} = \log x + C \)
\( \implies \) \( -\frac{1}{e^{y/x}} = \log x + C \)
\( \implies \) \( e^{y/x} \cdot \log x + C e^{y/x} + 1 = 0 \)
Putting \( x = 1, y = 1 \), we get
\( \therefore e \log 1 + C e + 1 = 0 \)
\( \implies \) \( C = -\frac{1}{e} \)
\( \therefore \) The required particular solution is
\( e^{y/x} \cdot \log x - \frac{1}{e} e^{y/x} + 1 = 0 \) or \( e^{y/x} \log x - e^{\frac{y}{x} - 1} + 1 = 0 \)

Question. Show that the differential equation \( \left[ x \sin^2\left(\frac{y}{x}\right) - y \right] dx + x dy = 0 \) is homogeneous. Find the particular solution of this differential equation, given that \( y = \frac{\pi}{4} \) when \( x = 1 \). 
Answer: Given differential equation is \( \left[ x \sin^2\left(\frac{y}{x}\right) - y \right] dx + x dy = 0 \)
\( \implies \) \( \frac{dy}{dx} = \frac{y - x \sin^2\left(\frac{y}{x}\right)}{x} \dots (i) \)
Let \( F(x, y) = \frac{y - x \sin^2\left(\frac{y}{x}\right)}{x} \)
Then \( F(\lambda x, \lambda y) = \frac{\lambda y - \lambda x \sin^2\left(\frac{\lambda y}{\lambda x}\right)}{\lambda x} = \lambda^0 \frac{y - x \sin^2\left(\frac{y}{x}\right)}{x} = \lambda^0 F(x, y) \)
Hence, differential equation (i) is homogeneous.
Now, let \( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Putting these value in (i), we get
\( v + x \frac{dv}{dx} = \frac{vx - x \sin^2 v}{x} \
\( \implies \) \( v + x \frac{dv}{dx} = \frac{x \{v - \sin^2 v\}}{x} \)
\( \implies \) \( v + x \frac{dv}{dx} = v - \sin^2 v \)
\( \implies \) \( x \frac{dv}{dx} = -\sin^2 v \)
\( \implies \) \( \frac{dv}{\sin^2 v} = -\frac{dx}{x} \)
Integrating both sides, we get
\( \implies \) \( \int \text{cosec}^2 v dv = -\int \frac{1}{x} dx \)
\( \implies \) \( -\cot v = -\log x + C \)
\( \implies \) \( \log x - \cot\left(\frac{y}{x}\right) = C \dots (ii) \)
Putting \( y = \frac{\pi}{4} \) and \( x = 1 \) in (ii), we get
\( \log 1 - \cot \frac{\pi}{4} = C \)
\( \implies \) \( 0 - 1 = C \)
\( \implies \) \( C = -1 \)
Hence, particular solution is
\( \log x - \cot\left(\frac{y}{x}\right) = -1 \)
\( \implies \) \( \log x - \cot\left(\frac{y}{x}\right) + 1 = 0 \)

Question. Find the differential equation of the family of curves \( (x - h)^2 + (y - k)^2 = r^2 \), where h and k are arbitrary constants.
Answer: Given family of curve is \( (x - h)^2 + (y - k)^2 = r^2 \). \dots (i)
Differentiating with respect to x, we get
\( \implies \) \( 2(x - h) + 2(y - k) \cdot \frac{dy}{dx} = 0 \)
\( \implies \) \( \frac{dy}{dx} = -\frac{x - h}{y - k} \dots (ii) \)
Differentiating again with respect to x, we get
\( \frac{d^2y}{dx^2} = -\left[ \frac{(y - k) - (x - h) \cdot \frac{dy}{dx}}{(y - k)^2} \right] = -\left[ \frac{(y - k) + (x - h) \cdot \frac{x - h}{y - k}}{(y - k)^2} \right] \) [From (ii)]
\( \implies \) \( \frac{d^2y}{dx^2} = -\left[ \frac{(y - k)^2 + (x - h)^2}{(y - k)^3} \right] = -\frac{r^2}{(y - k)^3} \) [From (i)] \dots (iii)
From (ii) \( \left( \frac{dy}{dx} \right)^2 = \left( \frac{x - h}{y - k} \right)^2 \)
\( \implies \) \( \left( \frac{dy}{dx} \right)^2 = \frac{(x - h)^2}{(y - k)^2} \)
Adding 1 both the sides, we get
\( \implies \) \( \left( \frac{dy}{dx} \right)^2 + 1 = \frac{(x - h)^2}{(y - k)^2} + 1 = \frac{(x - h)^2 + (y - k)^2}{(y - k)^2} \)
Putting exponent (power) \( \frac{3}{2} \) both sides, we get
\( \implies \) \( \left[ \left( \frac{dy}{dx} \right)^2 + 1 \right]^{3/2} = \left[ \frac{r^2}{(y - k)^2} \right]^{3/2} = \frac{r^3}{(y - k)^3} \)
\( \implies \) \( \left[ \left( \frac{dy}{dx} \right)^2 + 1 \right]^{3/2} = r \cdot \frac{r^2}{(y - k)^3} = -r \frac{d^2y}{dx^2} \) [Using (iii)]
\( \implies \) \( r \frac{d^2y}{dx^2} + \left[ \left( \frac{dy}{dx} \right)^2 + 1 \right]^{3/2} = 0 \)

Question. 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 differential equation is \( \frac{dy}{dx} = \frac{x(2 \log x + 1)}{\sin y + y \cos y} \)
\( \implies \) \( (\sin y + y \cos y) dy = x (2 \log x + 1) dx \)
\( \implies \) \( \int \sin y dy + \int y \cos y dy = 2 \int x \log x dx + \int x dx \)
\( \implies \) \( \int \sin y dy + \left[ y \sin y - \int \sin y dy \right] = 2 \left[ \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx \right] + \int x dx \)
\( \implies \) \( \int \sin y dy + y \sin y - \int \sin y dy = x^2 \log x - \int x dx + \int x dx + C \)
\( \implies \) \( y \sin y = x^2 \log x + C \), is general solution. \dots (i)
For particular solution, we put \( y = \frac{\pi}{2} \) when \( x = 1 \)
(i) becomes \( \frac{\pi}{2} \sin \frac{\pi}{2} = 1 \cdot \log 1 + C \)
\( \implies \) \( \frac{\pi}{2} = C \) \( [\because \log 1 = 0] \)
Putting the value of C in (i), we get the required particular solution
\( y \sin y = x^2 \log x + \frac{\pi}{2} \)

Question. Show that the family of curves for which the slope of the tangent at any point (x, y) on it is \( \frac{x^2 + y^2}{2xy} \), is given by \( x^2 - y^2 = Cx \). 
Answer: We know that the slope of the tangent at any point on a curve is \( \frac{dy}{dx} \).
Therefore, \( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \)
\( \implies \) \( \frac{dy}{dx} = \frac{1 + \frac{y^2}{x^2}}{\frac{2y}{x}} \dots (i) \)
Clearly, equation (i) is a homogeneous differential equation. To solve it we make substitution.
\( y = vx \)
\( \implies \) \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Putting the value of y and \( \frac{dy}{dx} \) in equation (i), we get
\( v + x \frac{dv}{dx} = \frac{1 + v^2}{2v} \)
\( \implies \) \( x \frac{dv}{dx} = \frac{1 - v^2}{2v} \)
\( \implies \) \( \frac{2v}{1 - v^2} dv = \frac{dx}{x} \)
\( \implies \) \( -\frac{2v}{v^2 - 1} dv = \frac{dx}{x} \)
Integrating both sides, we get
\( \int \frac{2v}{v^2 - 1} dv = -\int \frac{1}{x} dx \)
\( \implies \) \( \log|v^2 - 1| = -\log|x| + \log|C_1| \)
\( \implies \) \( \log|(v^2 - 1)x| = \log|C_1| \)
\( \implies \) \( (v^2 - 1)x = \pm C_1 \)
Replacing v by \( \frac{y}{x} \), we get
\( \left( \frac{y^2}{x^2} - 1 \right)x = \pm C_1 \)
\( \implies \) \( (y^2 - x^2) = \pm C_1 x \)
\( \implies \) \( x^2 - y^2 = Cx \) (where \( \pm C_1 = C \))