CBSE Class 12 Mathematics Differential Equations VBQs Set B

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

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

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
\(\frac{dy}{dx} = \frac{xy}{x^2+y^2}\) ...(i)
Now, to solve let \(y = vx\). [since (i) is a homogeneous equation]
Differentiating \(y = vx\) with respect to \(x\), we get
\(\frac{dy}{dx} = v + x \frac{dv}{dx}\)
Putting \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\) in (i), we get
\(v + x \frac{dv}{dx} = \frac{x \cdot vx}{x^2 + (vx)^2}\)
\(v + x \frac{dv}{dx} = \frac{vx^2}{x^2 + v^2x^2} \Rightarrow v + x \frac{dv}{dx} = \frac{vx^2}{x^2(1+v^2)}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{v}{(1+v^2)} - v \Rightarrow x \frac{dv}{dx} = \frac{v - v - v^3}{(1+v^2)}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{-v^3}{(1+v^2)} \Rightarrow \frac{(1+v^2) dv}{v^3} = -\frac{dx}{x}\)
Integrating both sides, we get
\(\int \frac{(1+v^2) dv}{v^3} = -\int \frac{dx}{x}\)
\(\Rightarrow \int \frac{dv}{v^3} + \int \frac{dv}{v} = -\log |x| + C \Rightarrow -\frac{1}{2v^2} + \log |v| = -\log |x| + C\)
\(\Rightarrow -\frac{x^2}{2y^2} + \log |\frac{y}{x}| = -\log |x| + C \Rightarrow -\frac{x^2}{2y^2} + \log |y| - \log |x| = -\log |x| + C\)
\(\Rightarrow -\frac{x^2}{2y^2} + \log |y| = C\) ...(ii)
Given, \(x = 1, y = 1\)
\(\Rightarrow -\frac{1}{2 \times 1} + \log |1| = C \Rightarrow -\frac{1}{2} = C\) [\(\because \log 1 = 0\)]
Now (ii) becomes
\(-\frac{x^2}{2y^2} + \log |y| = -\frac{1}{2} \Rightarrow \log |y| = \frac{x^2}{2y^2} - \frac{1}{2} \Rightarrow \log |y| = \frac{x^2 - y^2}{2y^2}\) ...(iii)
Putting \(x = x_0\) and \(y = e\) in (iii), we get
\(\log |e| = \frac{x_0^2 - e^2}{2e^2} \Rightarrow 1 = \frac{x_0^2 - e^2}{2e^2} \Rightarrow x_0^2 - e^2 = 2e^2\)
\(\Rightarrow x_0^2 = 3e^2 \Rightarrow x_0 = \sqrt{3}e\)

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, \(\frac{dy}{dx} + y \tan x = 3x^2 + x^3 \tan x\)
\(\Rightarrow \frac{dy}{dx} + \tan x \cdot y = 3x^2 + x^3 \tan x\)
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
\(y \cdot \sec x = \int (3x^2 + x^3 \tan x) \cdot \sec x dx + C\)
\(\Rightarrow y \cdot \sec x = \int 3x^2 \sec x dx + \int x^3 \tan x \sec x dx + C\)
\(\Rightarrow y \sec x = \int 3x^2 \sec x dx + x^3 \cdot \sec x - \int 3x^2 \cdot \sec x dx + C\)
\(\Rightarrow y \sec x = x^3 \sec x + C \Rightarrow y = x^3 + C \cos x\)
Now \(x = \frac{\pi}{3}, y = 0\)
\(\therefore 0 = \left(\frac{\pi}{3}\right)^3 + C \cdot \cos \left(\frac{\pi}{3}\right) \Rightarrow 0 = \frac{\pi^3}{27} + \frac{C}{2} \Rightarrow C = -\frac{2\pi^3}{27}\)
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. 
Answer: Given, \((x - y) \frac{dy}{dx} = x + 2y\)
By simplifying the above equation, we get
\(\frac{dy}{dx} = \frac{x + 2y}{x - y}\) ...(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 \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
Substituting these values in equation (i), we get
\(v + x \frac{dv}{dx} = \frac{x + 2vx}{x - vx}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v + v^2}{1 - v} = \frac{1 + v + v^2}{1 - v}\)
\(\Rightarrow \frac{1 - v}{1 + v + v^2} dv = \frac{dx}{x}\)
By integrating both sides, we get
\(\int \frac{1 - v}{1 + v + v^2} dv = \int \frac{dx}{x}\) ...(ii)
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 \(A = -\frac{1}{2}, B = \frac{3}{2}\)
\(\therefore \int \frac{1 - v}{v^2 + v + 1} dv = \int \frac{-\frac{1}{2}(2v + 1) + \frac{3}{2}}{v^2 + v + 1} dv\)
\(= -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} dv + \frac{3}{2} \int \frac{dv}{v^2 + v + 1}\)
\(= -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} dv + \frac{3}{2} \int \frac{dv}{(v + \frac{1}{2})^2 + \frac{3}{4}}\)
\(= -\frac{1}{2} \log |v^2 + v + 1| + \frac{3}{2} \times \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\)
Now, substituting it in equation (ii), we get
\(-\frac{1}{2} \log |v^2 + v + 1| + \sqrt{3} \tan^{-1} \left( \frac{2v + 1}{\sqrt{3}} \right) = \log x + C\)
\(\Rightarrow -\frac{1}{2} \log |\frac{y^2}{x^2} + \frac{y}{x} + 1| + \sqrt{3} \tan^{-1} \left( \frac{2\frac{y}{x} + 1}{\sqrt{3}} \right) = \log x + C\)
\(\Rightarrow -\frac{1}{2} \log |x^2 + xy + y^2| + \frac{1}{2} \log x^2 + \sqrt{3} \tan^{-1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log x + C\)
\(\Rightarrow -\frac{1}{2} \log |x^2 + xy + y^2| + \sqrt{3} \tan^{-1} \left( \frac{2y + x}{\sqrt{3}x} \right) = C\)

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

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 \(\frac{y-1}{x^2 + x}\).
\(\therefore \left(\frac{dy}{dx}\right)_{(x, y)} = \frac{y-1}{x^2 + x}\)
\(\Rightarrow \frac{dy}{dx} = \frac{y-1}{x^2 + x} \Rightarrow \frac{dy}{y-1} = \frac{dx}{x^2 + x}\)
On integrating both sides, we get \(\int \frac{dy}{y-1} = \int \frac{dx}{x^2 + x}\)
\(\Rightarrow \int \frac{dy}{y-1} = \int \frac{dx}{x(x + 1)} \Rightarrow \int \frac{dy}{y-1} = \int \left(\frac{1}{x} - \frac{1}{x + 1}\right) dx\)
\(\Rightarrow \log (y - 1) = \log x - \log (x + 1) + \log C\)
\(\Rightarrow \log (y - 1) = \log \left( \frac{xC}{x + 1} \right)\)
Since, the given curve passes through point (1, 0).
\(\therefore 0 - 1 = \frac{1 \cdot C}{1 + 1} \Rightarrow C = -2\)
The particular solution is \(y - 1 = \frac{-2x}{x + 1} \Rightarrow (y - 1)(x + 1) = -2x \Rightarrow (y - 1)(x + 1) + 2x = 0\).

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 \((1 - y^2)(1 + \log x) dx + 2xy dy = 0\)
\(\Rightarrow 2xy dy = -(1 - y^2)(1 + \log x) dx \Rightarrow \frac{2y}{1 - y^2} dy = -\frac{(1 + \log x) dx}{x}\)
Integrating both sides, we get
\(\int \frac{2y}{1 - y^2} dy = -\int \frac{(1 + \log x)}{x} dx \Rightarrow -\log |1 - y^2| = -\int \frac{(1 + \log x)}{x} dx\)
\(\Rightarrow -\log |1 - y^2| = -\int z dz\) [Let \(1 + \log x = z \Rightarrow \frac{1}{x} dx = dz\)]
\(\Rightarrow \log |1 - y^2| = \frac{z^2}{2} + C \Rightarrow \log |1 - y^2| = \frac{(1 + \log x)^2}{2} + C\)
Putting \(x = 1\) and \(y = 0\), we get
\(\Rightarrow \log 1 = \frac{(1 + \log 1)^2}{2} + C \Rightarrow 0 = \frac{1}{2} + C \Rightarrow C = -\frac{1}{2}\)
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 \((1 + y^2) + (x - e^{-\tan^{-1} y}) \frac{dy}{dx} = 0\)
\(\Rightarrow (x - e^{-\tan^{-1} y}) \frac{dy}{dx} = -(1 + y^2)\)
\(\Rightarrow \frac{dy}{dx} = -\frac{1 + y^2}{x - e^{-\tan^{-1} y}} \Rightarrow \frac{dx}{dy} = -\left( \frac{x - e^{-\tan^{-1} y}}{1 + y^2} \right)\)
\(\Rightarrow \frac{dx}{dy} = -\frac{x}{1 + y^2} + \frac{e^{-\tan^{-1} y}}{1 + y^2} \Rightarrow \frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{e^{-\tan^{-1} y}}{1 + y^2}\)
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 \(x \cdot e^{\tan^{-1} y} = \int \frac{e^{-\tan^{-1} y}}{1 + y^2} \cdot e^{\tan^{-1} y} dy + C\)
\(\Rightarrow x \cdot e^{\tan^{-1} y} = \int e^{-z} \cdot e^z dz + C\) [Let \(\tan^{-1} y = z \Rightarrow \frac{1}{1 + y^2} dy = dz\)]
\(\Rightarrow x \cdot e^{\tan^{-1} y} = \int e^0 dz + C \Rightarrow x \cdot e^{\tan^{-1} y} = z + C\)
\(\Rightarrow x \cdot e^{\tan^{-1} y} = \tan^{-1} y + C \Rightarrow x = (\tan^{-1} y) e^{-\tan^{-1} y} + C e^{-\tan^{-1} y}\)

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 \(\frac{dy}{dx} = - \frac{x + y \cos x}{1 + \sin x}\)
\(\Rightarrow \frac{dy}{dx} = -\frac{x}{1 + \sin x} - \frac{y \cos x}{1 + \sin x} \Rightarrow \frac{dy}{dx} + \frac{\cos x}{1 + \sin x} y = -\frac{x}{1 + \sin x}\)
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
\(y(1 + \sin x) = \int -\frac{x}{1 + \sin x}(1 + \sin x) dx + C = -\int x dx + C\)
\(\Rightarrow y(1 + \sin x) = -\frac{x^2}{2} + C\)
\(1(1 + \sin 0) = 0 + C \Rightarrow C = 1\) [Given \(y = 1\) and \(x = 0\)]
Hence, particular solution is
\(y(1 + \sin x) = -\frac{x^2}{2} + 1 \Rightarrow y = \frac{2 - x^2}{2(1 + \sin x)}\)

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
\(\Rightarrow \frac{dx}{dy} = \frac{\cot^{-1} y + x}{1 + y^2} = \frac{\cot^{-1} y}{1 + y^2} + \frac{x}{1 + y^2} \Rightarrow \frac{dx}{dy} - \frac{1}{1 + y^2} \cdot x = \frac{\cot^{-1} y}{1 + y^2}\)
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
\(x \cdot e^{\cot^{-1} y} = \int \frac{\cot^{-1} y}{1 + y^2} \cdot e^{\cot^{-1} y} dy + C \Rightarrow x \cdot e^{\cot^{-1} y} = I + C\) ... (i)
Here, \(I = \int \frac{\cot^{-1} y}{1 + y^2} \cdot e^{\cot^{-1} y} dy\)
Let \(\cot^{-1} y = t \Rightarrow -\frac{1}{1 + y^2} dy = dt \Rightarrow \frac{1}{1 + y^2} dy = -dt\)
\(\Rightarrow I = -\int t \cdot e^t dt = -[t \cdot e^t - \int e^t dt] = -t e^t + e^t = e^t(1 - t) = e^{\cot^{-1} y} (1 - \cot^{-1} y)\)
Hence, required solution is \(x e^{\cot^{-1} y} = e^{\cot^{-1} y} (1 - \cot^{-1} y) + C\). [From equation (i)]
\(\Rightarrow x = (1 - \cot^{-1} y) + C e^{-\cot^{-1} y}\)

Question. Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes. 
Answer: Let C denotes the family of circles in the second quadrant and touching the coordinate axes. Let \((-a, a)\) be the coordinate of the centre of any member of this family (see figure).
Equation representing the family C is
\((x + a)^2 + (y - a)^2 = a^2\) ...(i)
or \(x^2 + y^2 + 2ax - 2ay + a^2 = 0\) ...(ii)
Differentiating equation (ii) with respect to \(x\), we get
\(2x + 2y \frac{dy}{dx} + 2a - 2a \frac{dy}{dx} = 0 \Rightarrow x + y \frac{dy}{dx} = a \left( \frac{dy}{dx} - 1 \right)\)
or \(a = \frac{x + yy'}{y' - 1}\) (\(\because y' = \frac{dy}{dx}\))
Substituting the value of \(a\) in equation (i), we get
\(\left[x + \frac{x + yy'}{y' - 1}\right]^2 + \left[y - \frac{x + yy'}{y' - 1}\right]^2 = \left[\frac{x + yy'}{y' - 1}\right]^2\)
\(\Rightarrow [xy' - x + x + yy']^2 + [yy' - y - x - yy']^2 = [x + yy']^2\)
or \((x + y)^2 y'^2 + (x + y)^2 = (x + yy')^2 \Rightarrow (x + y)^2 [(y')^2 + 1] = [x + yy']^2\), is the required differential equation representing the given family of circles.

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 \(x \cos \left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos \left(\frac{y}{x}\right) + x\)
\(\frac{dy}{dx} = \frac{y \cos(y/x) + x}{x \cos(y/x)}\) ... (i)
It is homogeneous differential equation.
Let \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
(i) \(\Rightarrow v + x \frac{dv}{dx} = \frac{vx \cos v + x}{x \cos v}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{v \cos v + 1}{\cos v} - v \Rightarrow x \frac{dv}{dx} = \frac{v \cos v + 1 - v \cos v}{\cos v}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{1}{\cos v} \Rightarrow \cos v dv = \frac{dx}{x}\)
Integrating both sides
\(\Rightarrow \sin v = \log |x| + C \Rightarrow \sin \frac{y}{x} = \log |x| + C\) is the required solution.

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, \(3e^x \tan y dx + (2 - e^x) \sec^2 y dy = 0\)
\(\Rightarrow (2 - e^x) \sec^2 y dy = -3e^x \tan y dx\)
\(\frac{\sec^2 y}{\tan y} dy = \frac{-3e^x}{2 - e^x} dx \Rightarrow \int \frac{\sec^2 y}{\tan y} dy = 3 \int \frac{-e^x}{2 - e^x} dx\)
\(\Rightarrow \log |\tan y| = 3 \log |2 - e^x| + \log C\)
\(\Rightarrow \log |\tan y| = \log |C \cdot (2 - e^x)^3| \Rightarrow \tan y = C (2 - e^x)^3\)
Putting \(x = 0, y = \frac{\pi}{4}\), we get
\(\tan \frac{\pi}{4} = C(2 - e^0)^3 \Rightarrow 1 = C(2 - 1)^3 \Rightarrow 1 = C\)
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
\(v + x \frac{dv}{dx} = \frac{\sqrt{x^2 + v^2x^2} + vx}{x} \Rightarrow v + x \frac{dv}{dx} = \sqrt{1 + v^2} + v\)
\(\Rightarrow x \frac{dv}{dx} = \sqrt{1 + v^2} \Rightarrow \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\)
Integrating both sides, we get
\(\int \frac{1}{\sqrt{1 + v^2}} dv = \int \frac{1}{x} dx \Rightarrow \log |v + \sqrt{1 + v^2}| = \log |x| + \log C\)
\(\Rightarrow |v + \sqrt{1 + v^2}| = |Cx| \Rightarrow \left|\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}\right| = |Cx|\)
\(\Rightarrow \{y + \sqrt{x^2 + y^2}\}^2 = C^2x^4\) [Squaring both sides]
Hence, \(\{y + \sqrt{x^2 + y^2}\}^2 = C^2x^4\) 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 \((x e^{y/x} + y) dx = x dy \Rightarrow \frac{dy}{dx} = \frac{x e^{y/x} + y}{x}\) ... (i)
Let \(F(x, y) = \frac{x e^{y/x} + y}{x} \Rightarrow F(\lambda x, \lambda y) = \frac{\lambda x e^{\lambda y / \lambda x} + \lambda y}{\lambda x} = \lambda^0 \frac{x e^{y/x} + y}{x} = \lambda^0 F(x, y)\)
Hence, given differential equation (i) is homogeneous.
Let \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
Now, given differential equation (i) would become
\(v + x \frac{dv}{dx} = \frac{x e^v + vx}{x} \Rightarrow v + x \frac{dv}{dx} = e^v + v \Rightarrow x \frac{dv}{dx} = e^v\)
\(\Rightarrow \frac{dv}{e^v} = \frac{dx}{x} \Rightarrow \int e^{-v} dv = \int \frac{dv}{x} \Rightarrow \frac{e^{-v}}{-1} = \log x + C\)
\(\Rightarrow -e^{-y/x} = \log x + C \Rightarrow -\frac{1}{e^{y/x}} = \log x + C \Rightarrow e^{y/x} \cdot \log x + C e^{y/x} + 1 = 0\)
Putting \(x = 1, y = 1\), we get
\(\therefore e \log 1 + Ce + 1 = 0 \Rightarrow C = -\frac{1}{e}\)
\(\therefore\) The required particular solution is
\(e^{y/x} \cdot \log x - \frac{1}{e} e^{y/x} + 1 = 0\) or \(e^{y/x} \log x - e^{(y/x)-1} + 1 = 0\)

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 \(\left[x \sin^2\left(\frac{y}{x}\right) - y\right] dx + x dy = 0 \Rightarrow \frac{dy}{dx} = \frac{y - x \sin^2\left(\frac{y}{x}\right)}{x}\) ...(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 \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
Putting these values in (i), we get
\(v + x \frac{dv}{dx} = \frac{vx - x \sin^2\left(\frac{vx}{x}\right)}{x} \Rightarrow v + x \frac{dv}{dx} = \frac{x\{v - \sin^2 v\}}{x}\)
\(\Rightarrow v + x \frac{dv}{dx} = v - \sin^2 v \Rightarrow x \frac{dv}{dx} = -\sin^2 v \Rightarrow \frac{dv}{\sin^2 v} = -\frac{dx}{x}\)
Integrating both sides, we get
\(\int \text{cosec}^2 v dv = -\int \frac{1}{x} dx \Rightarrow -\cot v = -\log x + C \Rightarrow \log x - \cot\left(\frac{y}{x}\right) = C\) ...(ii)
Putting \(y = \frac{\pi}{4}\) and \(x = 1\) in (ii), we get
\(\log 1 - \cot \frac{\pi}{4} = C \Rightarrow 0 - 1 = C \Rightarrow C = -1\)
Hence, particular solution is
\(\log x - \cot\left(\frac{y}{x}\right) = -1 \Rightarrow \log x - \cot\left(\frac{y}{x}\right) + 1 = 0\)

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 \((x - h)^2 + (y - k)^2 = r^2\) ...(i)
Differentiating with respect to \(x\), we get
\(2(x - h) + 2(y - k) \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x - h}{y - k}\) ...(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)]
\(\Rightarrow \frac{d^2y}{dx^2} = -\left\{ \frac{(y - k)^2 + (x - h)^2}{(y - k)^3} \right\} = -\frac{r^2}{(y - k)^3}\) [From (i)] ...(iii)
From (ii) \(\left(\frac{dy}{dx}\right)^2 = \left(\frac{x - h}{y - k}\right)^2 \Rightarrow \left(\frac{dy}{dx}\right)^2 = \frac{(x - h)^2}{(y - k)^2}\)
Adding 1 both the sides, we get
\(\left(\frac{dy}{dx}\right)^2 + 1 = \frac{(x - h)^2}{(y - k)^2} + 1 = \frac{(x - h)^2 + (y - k)^2}{(y - k)^2}\)
Putting exponent (power) \(\frac{3}{2}\) both sides, we get
\(\left[ \left(\frac{dy}{dx}\right)^2 + 1 \right]^{3/2} = \left[ \frac{r^2}{(y - k)^2} \right]^{3/2} = \frac{r^3}{(y - k)^3}\)
\(\Rightarrow \left[ \left(\frac{dy}{dx}\right)^2 + 1 \right]^{3/2} = r \cdot \frac{r^2}{(y - k)^3} = -r \frac{d^2y}{dx^2}\) [Using (iii)]
\(\Rightarrow r \frac{d^2y}{dx^2} + \left[ \left(\frac{dy}{dx}\right)^2 + 1 \right]^{3/2} = 0\)

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 \(\frac{dy}{dx} = \frac{x(2 \log x + 1)}{\sin y + y \cos y}\)
\(\Rightarrow (\sin y + y \cos y) dy = x(2 \log x + 1) dx\)
\(\Rightarrow \int \sin y dy + \int y \cos y dy = 2 \int x \log x dx + \int x dx\)
\(\Rightarrow \int \sin y dy + [y \sin y - \int \sin y dy] = 2 \left[ \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx \right] + \int x dx\)
\(\Rightarrow \int \sin y dy + y \sin y - \int \sin y dy = x^2 \log x - \int x dx + \int x dx + C\)
\(\Rightarrow y \sin y = x^2 \log x + C\), is general solution. ...(i)
For particular solution, we put \(y = \frac{\pi}{2}\) when \(x = 1\)
(i) becomes \(\frac{\pi}{2} \sin \frac{\pi}{2} = 1 \cdot \log 1 + C \Rightarrow \frac{\pi}{2} = C\) [\(\because \log 1 = 0\)]
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, \(\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \Rightarrow \frac{dy}{dx} = \frac{1 + \frac{y^2}{x^2}}{\frac{2y}{x}}\) ...(i)
Clearly, equation (i) is a homogeneous differential equation. To solve it we make substitution.
\(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
Putting the value of \(y\) and \(\frac{dy}{dx}\) in equation (i), we get
\(v + x \frac{dv}{dx} = \frac{1 + v^2}{2v} \Rightarrow x \frac{dv}{dx} = \frac{1 - v^2}{2v}\)
\(\Rightarrow \frac{2v}{1 - v^2} dv = \frac{dx}{x} \Rightarrow \int \frac{2v}{v^2 - 1} dv = -\int \frac{dx}{x}\)
Integrating both sides, we get
\(\log |v^2 - 1| = -\log |x| + \log |C_1|\)
\(\Rightarrow \log |(v^2 - 1)(x)| = \log |C_1| \Rightarrow (v^2 - 1)x = \pm C_1\)
Replacing \(v\) by \(\frac{y}{x}\), we get
\(\left( \frac{y^2}{x^2} - 1 \right) x = \pm C_1 \Rightarrow (y^2 - x^2) = \pm C_1 x \Rightarrow x^2 - y^2 = Cx\) (where \(\pm C_1 = C\))

Objective Type Questions: 

Choose and write the correct option in each of the following questions.

Question. The degree of the differential equation \(\left(1 + \left(\frac{dy}{dx}\right)^3\right)^{2/3} = \left(\frac{d^2y}{dx^2}\right)^2\) is 
(a) 1
(b) 2
(c) 3
(d) 4
Answer: c

Question. The order and degree of the differential equation \(\frac{d^4y}{dx^4} = y + \left(\frac{dy}{dx}\right)^4\) are respectively
(a) 4, 1
(b) 4, 2
(c) 2, 2
(d) 2, 4
Answer: a

Question. The integrating factor of the differential equation \(x \frac{dy}{dx} - y = 2x^2\) is
(a) \(e^{-x}\)
(b) \(e^{-y}\)
(c) \(\frac{1}{x}\)
(d) \(x\)
Answer: c

Question. The differential equation of the family of lines passing through the origin is
(a) \(\frac{dy}{dx} = x\)
(b) \(\frac{dy}{dx} = y\)
(c) \(x \frac{dy}{dx} - y = 0\)
(d) \(x + \frac{dy}{dx} = 0\)
Answer: c

Question. Solution of the differential equation \(x \frac{dy}{dx} + y = x e^x\) is
(a) \(xy = e^x (1 - x) + C\)
(b) \(xy = e^x (x + 1) + C\)
(c) \(xy = e^y (y - 1) + C\)
(d) \(xy = e^x (x - 1) + C\)
Answer: d

Question. The general solution of the differential equation \(e^x dy + (y e^x + 2x) dx = 0\) is
(a) \(x e^y + x^2 = C\)
(b) \(x e^y + y^2 = C\)
(c) \(y e^x + x^2 = C\)
(d) \(y e^y + x^2 = C\)
Answer: c

Fill in the blanks.

Question. The number of arbitrary constant (s) in a particular solution of the differential equation \(\tan x dx + \tan y dy = 0\) is _____________ .
Answer: 0

Question. The solution of the differential equation \(x \frac{dy}{dx} + 2y = x^2\) is _____________ .
Answer: \(yx^2 = \frac{x^4}{4} + C\)

Question. The general solution of the differential equation \(\frac{dy}{dx} = \frac{x^2}{y^2}\) is _____________ .
Answer: \(y^3 - x^3 = C\)

Question. The degree of the differential equation \(\frac{d^2y}{dx^2} + e^{dy/dx} = 0\) is ____________ .
Answer: not defined