CBSE Class 12 Mathematics Differential Equations Assignment Set 07

Very Short Answer Type Questions (VSA)

Question. Find the order and the degree of the differential equation \( x^2 \frac{d^2y}{dx^2} = \left\{1 + \left(\frac{dy}{dx}\right)^2\right\}^4 \).
Answer: The given differential equation is \( x^2 \frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^4 \).
\(\therefore\) Its order is 2 and degree is 1.

Question. Write the sum of the order and degree of the differential equation \( \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + x^4 = 0 \).
Answer: Order = 2, Degree = 2.
\(\therefore\) Required Sum = \( 2 + 2 = 4 \)

Question. Write the degree of differential equation \( x^3 \left(\frac{d^2y}{dx^2}\right)^2 + x \left(\frac{dy}{dx}\right)^4 = 0 \).
Answer: Degree of the given differential equation is 2.

Question. Find the differential equation whose solution is \( v = \frac{A}{r} + B \), where \( A \) and \( B \) are arbitrary constants.
Answer: \( v = \frac{A}{r} + B \Rightarrow \frac{dv}{dr} = -\frac{A}{r^2} \Rightarrow \frac{d^2v}{dr^2} = \frac{2A}{r^3} \)
Now, \( \frac{d^2v}{dr^2} \div \frac{dv}{dr} = \frac{2A}{r^3} \div \left(-\frac{A}{r^2}\right) \Rightarrow \frac{d^2v}{dr^2} \div \frac{dv}{dr} = -\frac{2}{r} \Rightarrow \frac{d^2v}{dr^2} = -\frac{2}{r} \frac{dv}{dr} \)
\(\Rightarrow \frac{d^2v}{dr^2} + \frac{2}{r} \frac{dv}{dr} = 0 \) is the required differential equation.

Question. Find the differential equation whose solution is \( y = mx \), where \( m \) is an arbitrary constant.
Answer: Here, \( y = mx \) ...(i)
Differentiating (i) w.r.t. \( x \), we get \( \frac{dy}{dx} = m \) ...(ii)
Eliminating \( m \) from (i) and (ii), we get \( y = x \frac{dy}{dx} \Rightarrow x \frac{dy}{dx} - y = 0 \), is the required differential equation.

Question. Find the integrating factor of the differential equation \( (y + 3x^2) \frac{dx}{dy} = x \).
Answer: We have, \( (y + 3x^2) \frac{dx}{dy} = x \Rightarrow \frac{y + 3x^2}{x} = \frac{dy}{dx} \Rightarrow \frac{dy}{dx} - \frac{y}{x} = 3x \).
This is a linear differential equation.
\(\therefore \text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x)^{-1}} = \frac{1}{x} \)

Question. Find the integrating factor of the differential equation \( \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\right) \frac{dx}{dy} = 1 \).
Answer: We have, \( \left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\right) \frac{dx}{dy} = 1 \) or \( \frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \).
\(\therefore \text{I.F.} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2\sqrt{x}} \)

Question. Write the integrating factor of the differential equation \( (1 + x^2) + (2xy - \cot x) \frac{dx}{dy} = 0 \).
Answer: The given differential equation is \( (1 + x^2) + (2xy - \cot x) \frac{dx}{dy} = 0 \)
\(\Rightarrow (1 + x^2)\frac{dy}{dx} + 2xy - \cot x = 0 \Rightarrow \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2} \).
\(\therefore \text{I.F.} = e^{\int \frac{2x}{1+x^2} dx} = e^{\log(1+x^2)} = 1 + x^2 \).

Question. Write the solution of the differential equation \( \frac{dy}{dx} = 2^{-y} \).
Answer: We have, \( \frac{dy}{dx} = 2^{-y} \Rightarrow \frac{dy}{2^{-y}} = dx \Rightarrow 2^y dy = dx \) ...(i)
Integrating both sides of (i), we get \( \frac{2^y}{\log 2} = x + C \Rightarrow 2^y = (C + x)\log 2 \).
Taking log on both sides to the base 2, we get \( \log_2 2^y = \log_2 [(C + x)\log 2] \Rightarrow y = \log_2 [(C + x)\log 2] \), which is the required solution.

Question. Find the solution of the differential equation \( \frac{dy}{dx} = x^3 e^{-2y} \).
Answer: We have, \( \frac{dy}{dx} = x^3 e^{-2y} \Rightarrow e^{2y} dy = x^3 dx \).
On integrating, we get \( \frac{e^{2y}}{2} = \frac{x^4}{4} + C' \Rightarrow 2e^{2y} = x^4 + C \), where \( C = 4C' \).

Short Answer Type Questions (SA-I)

Question. Find the differential equation whose solution is \( y = e^{2x}(a + bx) \), where ‘a’ and ‘b’ are arbitrary constants.
Answer: We have \( y = e^{2x}(a + bx) \) ...(i)
On differentiating (i) w.r.t. \( x \), we get \( \frac{dy}{dx} = 2e^{2x}(a + bx) + be^{2x} = 2y + be^{2x} \) ...(ii)
Again differentiating (ii) w.r.t. \( x \), we get \( \frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2be^{2x} = 2\frac{dy}{dx} + 2\left(\frac{dy}{dx} - 2y\right) \) [Using (ii)]
\( = 4\frac{dy}{dx} - 4y \).
Hence, the required differential equation is \( \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0 \).

Question. Find the differential equation whose solution is \( y = a e^{bx+5} \), where \( a \) and \( b \) are arbitrary constants.
Answer: We have, \( y = a e^{bx+5} \) ...(i)
Differentiating (i) w.r.t. \( x \), we get \( \frac{dy}{dx} = a e^{bx+5} \cdot b \Rightarrow e^{bx+5} = \frac{1}{ab} \frac{dy}{dx} \) ...(ii)
Differentiating (ii) w.r.t. \( x \), we get \( b \cdot e^{bx+5} = \frac{1}{ab} \frac{d^2y}{dx^2} \Rightarrow e^{bx+5} = \frac{1}{ab^2} \frac{d^2y}{dx^2} \) ...(iii)
From (ii) & (iii), we have \( \frac{1}{ab} \frac{dy}{dx} = \frac{1}{ab^2} \frac{d^2y}{dx^2} \Rightarrow \frac{d^2y}{dx^2} = b \frac{dy}{dx} \) ...(iv)
From (i) and (ii), we have \( \frac{1}{y} \frac{dy}{dx} = b \) ...(v)
From (iv) and (v), we get \( \frac{d^2y}{dx^2} = \frac{1}{y} \left(\frac{dy}{dx}\right)^2 \) which is the required differential equation.

Question. Solve the differential equation \( \frac{dy}{dx} + y = \cos x - \sin x \).
Answer: We have, \( \frac{dy}{dx} + y = \cos x - \sin x \), which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = 1, Q = \cos x - \sin x \).
\(\therefore \text{I.F.} = e^{\int dx} = e^x \).
The solution of the given differential equation is \( ye^x = \int e^x (\cos x - \sin x) dx + C \)
\(\Rightarrow ye^x = e^x \cos x + C \Rightarrow y = \cos x + Ce^{-x} \).

Question. Find the general solution of the differential equation \( x e^{y/x} dy = (y e^{y/x} + x^2) dx \), \( x \neq 0 \).
Answer: We have, \( x e^{y/x} dy = (y e^{y/x} + x^2) dx \Rightarrow \frac{dy}{dx} = \frac{y}{x} + \frac{x}{e^{y/x}} \).
Putting \( \frac{y}{x} = t \Rightarrow y = xt \Rightarrow \frac{dy}{dx} = t + x \frac{dt}{dx} \).
\(\therefore\) The equation becomes, \( t + x \frac{dt}{dx} = t + \frac{x}{e^t} \Rightarrow x \frac{dt}{dx} = x e^{-t} \Rightarrow \frac{dt}{dx} = e^{-t} \Rightarrow dx = e^t dt \).
Integrating both sides, we get \( x = e^t + C \Rightarrow x = e^{y/x} + C \).

Question. Solve the differential equation \( \frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have, \( \frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 \Rightarrow \frac{dy}{dx} = (1 + x^2) + y^2 (1 + x^2) = (1 + x^2)(1 + y^2) \)
\(\Rightarrow \frac{dy}{1 + y^2} = (1 + x^2) dx \).
Integrating both sides, we get \( \tan^{-1} y = x + \frac{x^3}{3} + C \).
When \( x = 0, y = 1 \), we have \( \tan^{-1} 1 = 0 + 0 + C \Rightarrow C = \frac{\pi}{4} \).
\(\therefore \tan^{-1} y = x + \frac{x^3}{3} + \frac{\pi}{4} \) is the required solution.

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, \( (1 - y^2)(1 + \log x) dx + 2xy dy = 0 \Rightarrow (1 - y^2)(1 + \log x) dx = -2xy dy \)
\(\Rightarrow \frac{1 + \log x}{x} dx = -\frac{2y}{1 - y^2} dy \).
On integrating both sides, we get \( \frac{(1 + \log x)^2}{2} = \log |1 - y^2| + C \).
When \( x = 1, y = 0 \), we have \( \frac{(1 + \log 1)^2}{2} = \log |1| + C \Rightarrow C = \frac{1}{2} \).
\(\Rightarrow \frac{(1 + \log x)^2}{2} = \log |1 - y^2| + \frac{1}{2} \)
\(\Rightarrow (1 + \log x)^2 = 2\log |1 - y^2| + 1 \) is the required particular solution.

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: We have, \( \frac{dy}{dx} = \frac{x(2\log x + 1)}{\sin y + y \cos y} \Rightarrow (\sin y + y \cos y) dy = x (2\log x + 1) dx \).
On integrating both sides, we get \( -\cos y + y \sin y - (-\cos y) = \int x (2\log x + 1) dx \)
\(\Rightarrow y \sin y = 2 \left[ \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx \right] + \frac{x^2}{2} + C \)
\(\Rightarrow y \sin y = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C \Rightarrow y \sin y = x^2 \log x + C \).
When \( x = 1, y = \frac{\pi}{2} \), we have \( \frac{\pi}{2} \sin \frac{\pi}{2} = 1 \cdot \log(1) + C \Rightarrow C = \frac{\pi}{2} \).
\(\therefore y \sin y = x^2 \log x + \frac{\pi}{2} \) is the required particular solution.

Question. Find the particular solution of the differential equation \( x(1 + y^2) dx - y(1 + x^2) dy = 0 \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have, \( x(1 + y^2) dx - y(1 + x^2) dy = 0 \Rightarrow \frac{x}{1 + x^2} dx - \frac{y}{1 + y^2} dy = 0 \)
\(\Rightarrow \frac{2x}{1 + x^2} dx = \frac{2y}{1 + y^2} dy \).
Integrating both sides, we get \( \log(1 + y^2) = \log(1 + x^2) + \log C \Rightarrow 1 + y^2 = C(1 + x^2) \).
When \( x = 0, y = 1 \), we have \( 1 + 1 = C(1 + 0) \Rightarrow C = 2 \).
\(\therefore 1 + y^2 = 2(1 + x^2) \) is the required particular solution.

Question. Find the particular solution of the differential equation \( xy \frac{dy}{dx} = (x + 2)(y + 2) \); \( y = -1 \) when \( x = 1 \).
Answer: We have, \( xy \frac{dy}{dx} = (x + 2)(y + 2) \Rightarrow \frac{y dy}{y + 2} = \left(\frac{x + 2}{x}\right) dx \)
\(\Rightarrow \left(1 - \frac{2}{y + 2}\right) dy = \left(1 + \frac{2}{x}\right) dx \).
Integrating both sides, we get \( y - 2\log(y + 2) = x + 2\log x + C \).
When \( x = 1, y = -1 \), we have \( -1 - 2\log(-1 + 2) = 1 + 2\log 1 + C \Rightarrow C = -1 - 1 = -2 \).
So, we have \( y - 2\log(y + 2) = x + 2\log x - 2 \Rightarrow y - x + 2 = 2\log (x(y + 2)) \).

Question. Find the particular solution of the differential equation \( (x + 1) \frac{dy}{dx} = 2e^{-y} - 1 \); \( y = 0 \) when \( x = 0 \).
Answer: We have, \( (x + 1) \frac{dy}{dx} = 2e^{-y} - 1 \) ...(i)
\(\Rightarrow \frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1} \Rightarrow \frac{e^y}{2 - e^y} dy = \frac{dx}{x + 1} \).
Integrating both sides, we get \( -\log(2 - e^y) = \log(x + 1) + C \) ...(ii)
When \( x = 0, y = 0 \), we have \( -\log(2 - 1) = \log(0 + 1) + C \Rightarrow C = 0 \).
\(\therefore\) (ii) becomes \( -\log(2 - e^y) = \log(x + 1) \Rightarrow \log(x + 1) + \log(2 - e^y) = 0 \)
\(\Rightarrow \log[(x + 1)(2 - e^y)] = 0 \Rightarrow (x + 1)(2 - e^y) = 1 \) is the required particular solution.

Short Answer Type Questions (SA-II)

Question. Find the differential equation whose solution is \( (x + a)^2 + (y - a)^2 = a^2 \), where \( a \) is constant.
Answer: We have \( (x + a)^2 + (y - a)^2 = a^2 \) ...(i) which has only one arbitrary constant \( a \).
Differentiating (i) w.r.t. \( x \), we get \( 2(x + a) + 2(y - a)\frac{dy}{dx} = 0 \Rightarrow a = -\frac{x + yy'}{y' - 1} \) ...(ii)
Substituting value of \( a \) from (ii) in (i), we get
\( \left(x + \frac{x + yy'}{y' - 1}\right)^2 + \left(y - \frac{x + yy'}{y' - 1}\right)^2 = \left(\frac{x + yy'}{y' - 1}\right)^2 \)
\(\Rightarrow [x(y' - 1) + x + y'y]^2 + [y(y' - 1) - x - y'y]^2 = (x + y'y)^2 \)
\(\Rightarrow (x + y)^2 (y')^2 + (x + y)^2 = (x + y'y)^2 \)
\(\Rightarrow (x + y)^2 \left[\left(\frac{dy}{dx}\right)^2 + 1\right] = \left[x + y \frac{dy}{dx}\right]^2 \), is the required differential equation.

Question. Find the differential equation whose solution is \( x^2 = 4ay \), where \( a \) is constant.
Answer: We have \( x^2 = 4ay \), where \( a \) is the constant. ...(i)
Differentiating (i) w.r.t. \( x \), we get \( 2x = 4ay_1 \Rightarrow \frac{2x}{y_1} = 4a \) ...(ii)
Substituting the value of \( 4a \) from (ii) in (i), we get \( x^2 = \frac{2x}{y_1} y \Rightarrow x^2 y_1 - 2xy = 0 \Rightarrow xy_1 - 2y = 0 \)
\(\Rightarrow x \frac{dy}{dx} - 2y = 0 \), is the required differential equation.

Question. Find the general solution of the differential equation \( x e^x dy = (x^3 + 2y e^x) dx \).
Answer: We have, \( x e^x dy = (x^3 + 2y e^x) dx \Rightarrow \frac{dy}{dx} = \frac{x^3 + 2y e^x}{x e^x} \Rightarrow \frac{dy}{dx} - \frac{2}{x} y = x^2 e^{-x} \) ...(i)
This is a linear D.E. of the form \( \frac{dy}{dx} + Py = Q \).
\(\therefore \text{I.F.} = e^{\int -\frac{2}{x} dx} = e^{-2\log x} = e^{\log x^{-2}} = \frac{1}{x^2} \).
So, the solution of (i) is \( y \cdot \frac{1}{x^2} = \int \frac{1}{x^2} \cdot x^2 e^{-x} dx \Rightarrow \frac{y}{x^2} = -e^{-x} + C \Rightarrow y = -x^2 e^{-x} + C x^2 \) which is the required solution.

Question. Find the particular solution of the differential equation \( x \frac{dy}{dx} = y - x \tan\left(\frac{y}{x}\right) \), given that \( y = \frac{\pi}{4} \) at \( x = 1 \).
Answer: We have, \( x \frac{dy}{dx} = y - x \tan\left(\frac{y}{x}\right) \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \tan\left(\frac{y}{x}\right) \), which is a homogeneous differential equation.
Now, put \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\(\therefore v + x \frac{dv}{dx} = v - \tan v \Rightarrow x \frac{dv}{dx} = -\tan v \Rightarrow \frac{dv}{\tan v} = -\frac{dx}{x} \Rightarrow \cot v dv + \frac{dx}{x} = 0 \).
Integrating both sides, we get \( \log|\sin v| + \log x = \log C \Rightarrow x \sin v = C \Rightarrow x \sin\left(\frac{y}{x}\right) = C \).
When \( x = 1, y = \frac{\pi}{4} \), we have \( 1 \cdot \sin\left(\frac{\pi}{4}\right) = C \Rightarrow C = \frac{1}{\sqrt{2}} \).
So, \( x \sin\left(\frac{y}{x}\right) = \frac{1}{\sqrt{2}} \) is the required particular solution.

Question. Solve the differential equation \( x dy - y dx = \sqrt{x^2 + y^2} dx \), given that \( y = 0 \) when \( x = 1 \).
Answer: We have, \( x \frac{dy}{dx} - y = \sqrt{x^2 + y^2} \Rightarrow \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} \) ...(i)
This is a homogeneous differential equation.
Put \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\(\therefore\) (i) becomes \( v + x \frac{dv}{dx} = v + \sqrt{1 + v^2} \Rightarrow x \frac{dv}{dx} = \sqrt{1 + v^2} \Rightarrow \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \)
\(\Rightarrow \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} \Rightarrow \log|v + \sqrt{1 + v^2}| = \log x + \log C_1 \)
\(\Rightarrow \log C_1 x = \log \left| \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} \right| \Rightarrow \log C_1 x = \log \left| y + \sqrt{x^2 + y^2} \right| - \log x \)
\(\Rightarrow \pm C_1 x^2 = y + \sqrt{x^2 + y^2} \Rightarrow C x^2 = y + \sqrt{x^2 + y^2} \) [where \( C = \pm C_1 \)].
When \( x = 1, y = 0 \), we have \( C \cdot 1 = 0 + \sqrt{1 + 0} \Rightarrow C = 1 \).
\(\therefore\) Required particular solution is \( x^2 = y + \sqrt{x^2 + y^2} \).

Question. Solve the differential equation \( (1 + x^2) \frac{dy}{dx} + 2xy - 4x^2 = 0 \), given \( y(0) = 0 \).
Answer: We have \( (1 + x^2)\frac{dy}{dx} + 2xy = 4x^2 \Rightarrow \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{4x^2}{1 + x^2} \).
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2x}{1 + x^2} \) and \( Q = \frac{4x^2}{1 + x^2} \).
\(\therefore \text{I.F.} = e^{\int Pdx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\log(1 + x^2)} = 1 + x^2 \).
Hence, the required solution is \( y(1 + x^2) = \int \frac{4x^2}{1 + x^2} (1 + x^2) dx + C \Rightarrow y(1 + x^2) = \int 4x^2 dx + C \)
\(\Rightarrow y(1 + x^2) = \frac{4x^3}{3} + C \).
Given that \( y(0) = 0 \Rightarrow 0(1 + 0) = 0 + C \Rightarrow C = 0 \).
Thus, \( y = \frac{4x^3}{3(1 + x^2)} \) is the required solution.

Question. Find the particular solution of the differential equation \( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have, \( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \).
This is a homogeneous differential equation. \(\therefore\) Put \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\(\therefore v + x \frac{dv}{dx} = \frac{x \cdot vx}{x^2 + v^2 x^2} \Rightarrow v + x \frac{dv}{dx} = \frac{v}{1 + v^2} \Rightarrow x \frac{dv}{dx} = \frac{v}{1 + v^2} - v \)
\(\Rightarrow x \frac{dv}{dx} = \frac{-v^3}{1 + v^2} \Rightarrow \frac{1 + v^2}{v^3} dv = -\frac{dx}{x} \).
Integrating both sides, we get \( \int \frac{dx}{x} = -\int v^{-3} dv - \int \frac{1}{v} dv \Rightarrow \log x = \frac{1}{2v^2} - \log v + C \)
\(\Rightarrow \log x = \frac{x^2}{2y^2} - \log y + \log x + C \Rightarrow \log y = \frac{x^2}{2y^2} + C \).
When \( x = 0, y = 1 \Rightarrow \log 1 = 0 + C \Rightarrow C = 0 \).
\(\therefore\) Particular solution is \( \log y = \frac{x^2}{2y^2} \Rightarrow y = e^{\frac{x^2}{2y^2}} \).

Question. Find the particular solution of the differential equation \( e^x \tan y dx + (2 - e^x)\sec^2 y dy = 0 \), given that \( y = \frac{\pi}{4} \) when \( x = 0 \).
Answer: The given differential equation is, \( e^x \tan y dx + (2 - e^x) \sec^2 y dy = 0 \)
\(\Rightarrow (2 - e^x)\sec^2 y dy = -e^x \tan y dx \Rightarrow \frac{\sec^2 y}{\tan y} dy = \frac{-e^x}{2 - e^x} dx \).
Integrating both sides, we get \( \int \frac{\sec^2 y}{\tan y} dy = \int \frac{-e^x}{2 - e^x} dx \Rightarrow \log \tan y = \log(2 - e^x) + C \).
When \( x = 0, y = \frac{\pi}{4} \), we have \( \log \tan \frac{\pi}{4} = \log(2 - e^0) + C \Rightarrow 0 = \log 1 + C \Rightarrow C = 0 \).
\(\therefore\) Particular solution is \( \log \tan y = \log(2 - e^x) \), i.e., \( e^x + \tan y - 2 = 0 \).

Question. Solve the following differential equation \( y^2 dx + (x^2 - xy + y^2) dy = 0 \).
Answer: We have, \( y^2 dx + (x^2 - xy + y^2) dy = 0 \Rightarrow \frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2} \).
This is a homogeneous differential equation. \(\therefore\) Put \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \), we get
\( v + x \frac{dv}{dx} = \frac{-v^2 x^2}{x^2 - v x^2 + v^2 x^2} \Rightarrow v + x \frac{dv}{dx} = \frac{-v^2}{1 - v + v^2} \)
\(\Rightarrow x \frac{dv}{dx} = \frac{-v^2}{1 - v + v^2} - v \Rightarrow x \frac{dv}{dx} = \frac{-v - v^3}{1 - v + v^2} \Rightarrow \frac{1 - v + v^2}{v(1 + v^2)} dv = -\frac{1}{x} dx \).
Integrating both sides, we get \( \int \frac{1 + v^2}{v(1 + v^2)} dv - \int \frac{v}{v(1 + v^2)} dv = -\int \frac{1}{x} dx \)
\(\Rightarrow \int \frac{1}{v} dv - \int \frac{1}{1 + v^2} dv = -\int \frac{1}{x} dx \Rightarrow \log|v| - \tan^{-1} v = -\log|x| + \log C \)
\(\Rightarrow \log\left| \frac{vx}{C} \right| = \tan^{-1} v \Rightarrow \left| \frac{vx}{C} \right| = e^{\tan^{-1} v} \Rightarrow |y| = C e^{\tan^{-1}(y/x)} \) is the required solution.

Question. Solve the differential equation \( (x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1} \), \( |x| \neq 1 \).
Answer: We have, \( (x^2 - 1)\frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1} \Rightarrow \frac{dy}{dx} + \frac{2x}{x^2 - 1} y = \frac{2}{(x^2 - 1)^2} \).
This is a linear differential equation of the form, \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{2x}{x^2 - 1} \) and \( Q = \frac{2}{(x^2 - 1)^2} \).
\(\therefore \text{I.F.} = e^{\int Pdx} = e^{\int \frac{2x}{x^2 - 1} dx} = e^{\log(x^2 - 1)} = x^2 - 1 \).
Hence, solution of the differential equation is given by \( y(x^2 - 1) = \int \frac{2(x^2 - 1)}{(x^2 - 1)^2} dx \)
\(\Rightarrow y(x^2 - 1) = 2\int \frac{dx}{x^2 - 1} \Rightarrow y(x^2 - 1) = 2 \times \frac{1}{2} \log\left| \frac{x - 1}{x + 1} \right| + C \)
\(\Rightarrow y(x^2 - 1) = \log\left| \frac{x - 1}{x + 1} \right| + C \).

Question. Find the particular solution of the differential equation \( \frac{dy}{dx} = 1 + x + y + xy \), given that \( y = 0 \) when \( x = 1 \).
Answer: We have, \( \frac{dy}{dx} = 1 + x + y + xy \Rightarrow \frac{dy}{dx} = (1 + x) + (1 + x)y = (1 + x)(1 + y) \)
\(\Rightarrow \frac{dy}{1 + y} = (1 + x) dx \).
Integrating both sides, we get \( \int \frac{dy}{1 + y} = \int (1 + x) dx + C \Rightarrow \log(1 + y) = x + \frac{x^2}{2} + C \) ...(i)
When \( x = 1, y = 0 \), we have \( \log 1 = 1 + \frac{1}{2} + C \Rightarrow C = -\frac{3}{2} \).
\(\therefore\) The particular solution is \( \log(1 + y) = x + \frac{x^2}{2} - \frac{3}{2} \).

Question. Solve the differential equation \( \frac{dy}{dx} + y \cot x = 2\cos x \), given that \( y = 0 \) when \( x = \frac{\pi}{2} \).
Answer: We have, \( \frac{dy}{dx} + y \cot x = 2\cos x \).
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \cot x \), \( Q = 2\cos x \).
\(\therefore \text{I.F.} = e^{\int \cot x dx} = e^{\log|\sin x|} = \sin x \).
\(\therefore y|\sin x| = \int |\sin x| (2\cos x) dx \Rightarrow y \sin x = \int \sin 2x dx \Rightarrow y\sin x = -\frac{1}{2}\cos 2x + C \).
When \( x = \frac{\pi}{2}, y = 0 \), we have \( 0\left(\sin \frac{\pi}{2}\right) = -\frac{1}{2}\cos\left(2\frac{\pi}{2}\right) + C \Rightarrow C = -\frac{1}{2} \).
\(\therefore y(\sin x) = -\frac{1}{2}\cos 2x - \frac{1}{2} \), i.e., \( 2y\sin x + \cos 2x + 1 = 0 \) is the required solution.

Question. If \( y(x) \) is a solution of the differential equation \( \left(\frac{2 + \sin x}{1 + y}\right) \frac{dy}{dx} = -\cos x \) and \( y(0) = 1 \), then find the value of \( y\left(\frac{\pi}{2}\right) \).
Answer: We have, \( \left(\frac{2 + \sin x}{1 + y}\right) \frac{dy}{dx} = -\cos x \Rightarrow \frac{dy}{1 + y} = -\frac{\cos x}{2 + \sin x} dx \).
Integrating both sides, we get \( \log(y + 1) = -\log|2 + \sin x| + \log C \Rightarrow \log(y + 1) = \log\left(\frac{C}{2 + \sin x}\right) \)
\(\Rightarrow y + 1 = \frac{C}{2 + \sin x} \Rightarrow (y + 1)(2 + \sin x) = C \).
Given: \( y(0) = 1 \Rightarrow x = 0, y = 1 \).
\(\therefore (1 + 1)(2 + \sin 0) = C \Rightarrow C = 4 \).
\(\therefore (y + 1)(2 + \sin x) = 4 \Rightarrow y = \frac{4}{2 + \sin x} - 1 \) ...(i)
Put \( x = \frac{\pi}{2} \) in (i), we get \( y\left(\frac{\pi}{2}\right) = \frac{4}{2 + 1} - 1 = \frac{1}{3} \).

Question. Find the particular solution of the following differential equation \( x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0 \), given that when \( x = 2 \), \( y = \pi \).
Answer: We have \( x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0 \Rightarrow \frac{dy}{dx} - \frac{y}{x} + \sin\left(\frac{y}{x}\right) = 0 \).
This is a linear homogeneous differential equation. Put \( y = vx \Rightarrow \frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} \).
\(\therefore v + x \frac{dv}{dx} - v + \sin v = 0 \Rightarrow x \frac{dv}{dx} + \sin v = 0 \Rightarrow \text{cosec } v dv + \frac{dx}{x} = 0 \).
Integrating both sides, we get \( \log|\text{cosec } v - \cot v| + \log x = \log C \Rightarrow x(\text{cosec } v - \cot v) = C \)
\(\Rightarrow x \left[ \text{cosec}\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) \right] = C \).
When \( x = 2, y = \pi \), we have \( 2 \left[ \text{cosec } \frac{\pi}{2} - \cot \frac{\pi}{2} \right] = C \Rightarrow C = 2 \).
\(\Rightarrow x \left[ \text{cosec}\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) \right] = 2 \) is the required particular solution.

Question. Show that the differential equation \( \frac{dy}{dx} = \frac{y^2}{xy - x^2} \) is homogeneous and also solve it.
Answer: We have, \( \frac{dy}{dx} = \frac{y^2}{xy - x^2} = \frac{y^2/x^2}{(xy - x^2)/x^2} \) ...(i)
\(\Rightarrow \frac{dy}{dx} = \frac{y^2/x^2}{\frac{y}{x} - 1} \).
\(\therefore\) It is a homogeneous differential equation.
Put \( y = vx \Rightarrow \frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} \).
\(\therefore\) (i) becomes \( v + x \frac{dv}{dx} = \frac{v^2}{v - 1} \Rightarrow x \frac{dv}{dx} = \frac{v^2}{v - 1} - v \Rightarrow x \frac{dv}{dx} = \frac{v}{v - 1} \)
\(\Rightarrow \frac{v - 1}{v} dv = \frac{dx}{x} \Rightarrow \left(1 - \frac{1}{v}\right) dv = \frac{dx}{x} \).
Integrating both sides, we get \( v - \log v = \log x + C \Rightarrow v = \log vx + C \)
\(\Rightarrow \frac{y}{x} = \log y + C \Rightarrow y = x(\log y + C) \) is the required solution.

Long Answer Type Questions (LA)

Question. Find the particular solution of the differential equation \( (x - y) \frac{dy}{dx} = x + 2y \), given that \( y = 0 \) when \( x = 1 \).
Answer: We have, \( (x - y) \frac{dy}{dx} = x + 2y \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y} \) ...(i)
Put \( y = Vx \Rightarrow \frac{dy}{dx} = V + x \frac{dV}{dx} \).
Putting \( \frac{dy}{dx} = V + x \frac{dV}{dx} \) in (i), we get \( V + x \frac{dV}{dx} = \frac{x + 2Vx}{x - Vx} \Rightarrow V + x \frac{dV}{dx} = \frac{1 + 2V}{1 - V} \)
\(\Rightarrow x \frac{dV}{dx} = \frac{1 + 2V}{1 - V} - V \Rightarrow x \frac{dV}{dx} = \frac{1 + 2V - V + V^2}{1 - V} \)
\(\Rightarrow \int \frac{1 - V}{V^2 + V + 1} dV = \int \frac{1}{x} dx \Rightarrow \int \frac{2 - 2V}{V^2 + V + 1} dV = 2\log|x| + c \)
\(\Rightarrow \int \frac{3 - (2V + 1)}{V^2 + V + 1} dV = 2\log|x| + c \)
\(\Rightarrow \int \frac{3}{V^2 + V + 1} dV - \int \frac{2V + 1}{V^2 + V + 1} dV = \log|x^2| + c \)
\(\Rightarrow 3\int \frac{1}{\left(V + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dV - \log|V^2 + V + 1| = \log |x^2| + c \)
\(\Rightarrow 3 \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{V + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \log|x^2 (V^2 + V + 1)| + c \)
\(\Rightarrow 2\sqrt{3} \tan^{-1}\left(\frac{2V + 1}{\sqrt{3}}\right) = \log|x^2 (V^2 + V + 1)| + c \).
Substituting \( V = \frac{y}{x} \) in above, we get \( 2\sqrt{3} \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \log \left| x^2 \frac{(y^2 + yx + x^2)}{x^2} \right| + c \)
\(\Rightarrow 2\sqrt{3} \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \log|y^2 + xy + x^2| + c \).
Now, at \( y = 0 \) and \( x = 1 \), we have \( 2\sqrt{3} \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \log|1| + c \Rightarrow c = 2\sqrt{3} \cdot \frac{\pi}{6} = \frac{\pi}{\sqrt{3}} \).
Substituting \( c = \frac{\pi}{\sqrt{3}} \) in above, we get \( 2\sqrt{3} \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \log|y^2 + xy + x^2| + \frac{\pi}{\sqrt{3}} \)
\(\Rightarrow 6 \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \sqrt{3}\log(y^2 + xy + x^2) + \pi \).

Question. Find the particular solution of the differential equation \( (\tan^{-1}x - y) dx = (1 + x^2) dy \), given that \( y = 1 \) when \( x = 0 \).
Answer: We have, \( (\tan^{-1}x - y)dx = (1 + x^2)dy \Rightarrow \frac{dy}{dx} = \frac{\tan^{-1}x - y}{1 + x^2} \Rightarrow \frac{dy}{dx} + \frac{1}{1 + x^2} \cdot y = \frac{\tan^{-1}x}{1 + x^2} \).
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \frac{1}{1 + x^2} \) and \( Q = \frac{\tan^{-1}x}{1 + x^2} \).
\(\therefore \text{I.F.} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x} \).
\(\therefore\) Solution is given by \( y e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \left( \frac{\tan^{-1}x}{1 + x^2} \right) dx + C \)
\(\Rightarrow y e^{\tan^{-1}x} = e^{\tan^{-1}x} \cdot \tan^{-1}x - \int e^{\tan^{-1}x} \cdot 1 \cdot \frac{1}{1+x^2} dx \cdot (1+x^2) + C \)
\(\Rightarrow y = \tan^{-1} x - 1 + C e^{-\tan^{-1}x} \) ...(i)
Now, putting \( x = 0, y = 1 \) in (i), we get \( 1 = \tan^{-1} 0 - 1 + C e^{-\tan^{-1}(0)} \Rightarrow C = 2 \).
So, required particular solution is \( y = \tan^{-1} x - 1 + 2e^{-\tan^{-1}x} \).

Question. Solve the following differential equation: \( \left(\sqrt{1 + x^2 + y^2 + x^2 y^2}\right) dx + xy dy = 0 \).
Answer: We have, \( \sqrt{1 + x^2 + y^2 + x^2y^2} dx + xy dy = 0 \Rightarrow \frac{dy}{dx} = -\frac{\sqrt{(1 + x^2)(1 + y^2)}}{xy} \)
\(\Rightarrow \frac{y}{\sqrt{1 + y^2}} dy = -\frac{\sqrt{1 + x^2}}{x^2} x dx \Rightarrow \int \frac{2y}{2\sqrt{1 + y^2}} dy = -\int \frac{\sqrt{1+x^2}}{x^2} x dx \).
[putting \( 1 + x^2 = v^2 \Rightarrow 2xdx = 2vdv \)]
\(\Rightarrow \sqrt{1 + y^2} = -\int \left( 1 + \frac{1}{v^2 - 1} \right) dv \Rightarrow \sqrt{1 + y^2} = -v - \frac{1}{2} \log \left| \frac{v - 1}{v + 1} \right| + C \)
\(\Rightarrow \sqrt{1 + y^2} + \sqrt{1 + x^2} + \frac{1}{2} \log \left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right| = C \).

Question. Find the particular solution of the differential equation \( (x - y) \frac{dy}{dx} = x + 2y \), given that when \( x = 1 \), \( y = 0 \).
Answer: We have, \( (x - y) \frac{dy}{dx} = x + 2y \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y} \) ...(i)
This is a linear homogeneous differential equation. \(\therefore\) Put \( y = vx \Rightarrow \frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} \).
\(\therefore\) (i) becomes \( v + x \frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v} \)
\(\Rightarrow x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + v + v^2}{1 - v} \Rightarrow \frac{1 - v}{1 + v + v^2} dv = \frac{dx}{x} \).
Integrating both sides, we get \( -\frac{1}{2} \int \frac{(2v + 1) - 3}{v^2 + v + 1} dv = \log x + C \)
\(\Rightarrow -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} dv + \frac{3}{2} \int \frac{dv}{\left(v + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \log x + C \)
\(\Rightarrow -\frac{1}{2} \log(v^2 + v + 1) + \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = \log x + C \).
The general solution is \( \log x + C = -\frac{1}{2} \log\left(\frac{y^2}{x^2} + \frac{y}{x} + 1\right) + \sqrt{3} \tan^{-1} \left[ \left(\frac{2y}{x} + 1\right)/\sqrt{3} \right] \) ...(i)
Putting \( x = 1, y = 0 \) in (i), we get \( 0 + C = -\frac{1}{2}\log(0 + 0 + 1) + \sqrt{3} \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \Rightarrow C = \frac{\pi}{2\sqrt{3}} \).
\(\therefore \log x + \frac{\pi}{2\sqrt{3}} = -\frac{1}{2} [\log(y^2 + xy + x^2) - \log x^2] + \sqrt{3} \tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) \)
\(\Rightarrow \frac{\pi}{2\sqrt{3}} = -\frac{1}{2}\log(x^2 + xy + y^2) + \sqrt{3}\tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) \).

Question. Prove that \( x^2 - y^2 = C(x^2 + y^2)^2 \) is the general solution of the differential equation \( (x^3 - 3xy^2)dx = (y^3 - 3x^2y) dy \), where \( C \) is a parameter.
Answer: We have, \( (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \Rightarrow \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \) ...(i)
Put, \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\(\therefore\) (i) becomes \( v + x \frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} \Rightarrow x \frac{dv}{dx} = \frac{1 - 3v^2 - v^4 + 3v^2}{v^3 - 3v} \Rightarrow x \frac{dv}{dx} = \frac{1 - v^4}{v(v^2 - 3)} \)
\(\Rightarrow \frac{v(v^2 - 3)dv}{1 - v^4} = \frac{dx}{x} \Rightarrow \int \frac{(v^3 - 3v)dv}{(1 - v^2)(1 + v^2)} = \int \frac{dx}{x} \) ...(ii)
Now, let \( \frac{v^3 - 3v}{(1 - v^2)(1 + v^2)} = \frac{Av + B}{1 - v^2} + \frac{Cv + D}{1 + v^2} \) ...(iii)
\(\Rightarrow v^3 - 3v = (Av + B)(1 + v^2) + (Cv + D)(1 - v^2) \).
Comparing coeff. of like powers, we get \( A - C = 1, A + C = -3, B - D = 0 \) and \( B + D = 0 \).
Solving these equations, we get \( A = -1, B = 0, C = -2, D = 0 \) ...(iv)
From (ii), (iii) and (iv), we have \( \int \frac{-v}{1 - v^2} dv - \int \frac{2v}{1 + v^2} dv = \int \frac{dx}{x} \)
\(\Rightarrow \frac{1}{2} \log(1 - v^2) - \log(1 + v^2) = \log x + \log C_1 \)
\(\Rightarrow \frac{\sqrt{1 - v^2}}{1 + v^2} = C_1 x \Rightarrow \frac{\sqrt{x^2 - y^2}}{x^2 + y^2} = C_1 x \)
\(\Rightarrow x^2 - y^2 = C_1^2 (x^2 + y^2)^2 \)
i.e., \( x^2 - y^2 = C(x^2 + y^2)^2 \) (where \( C_1^2 = C \)) which is the required solution. Hence Proved.