CBSE Class 12 Mathematics Differential Equations Worksheet Set 03

Selected NCERT Questions

Question. \( xy = \log y + C : y' = \frac{y^2}{1 - xy} (xy \neq -1) \).
Answer: Sol. The given function is \( xy = \log y + C \). Now,
\( x \cdot \frac{dy}{dx} + y = \frac{1}{y} \frac{dy}{dx} \)
\( \implies \left( \frac{1}{y} - x \right) \frac{dy}{dx} = y \)
\( \implies \frac{dy}{dx} = \frac{y^2}{1 - xy} \)
which is given differential equation.
Thus, \( xy = \log y + C \) is a solution of the given differential equation.

Question. \( y - \cos y = x : (y \sin y + \cos y + x) y' = y \).
Answer: Sol. The given function is \( y - \cos y = x \).
Now, \( \frac{dy}{dx} + \sin y \frac{dy}{dx} = 1 \)
\( \implies \frac{dy}{dx} = \frac{1}{1 + \sin y} \)
Putting values of \( y' \) and \( x \) in given differential equation, we have
LHS \( = (y \sin y + \cos y + y - \cos y) \cdot \frac{1}{(1 + \sin y)} = y (1 + \sin y) \cdot \frac{1}{(1 + \sin y)} = y = \text{RHS} \)
Thus, \( y - \cos y = x \) is a solution of the given differential equation.

Question. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Answer: Sol. We know that the differential equation of the family of hyperbolas having foci on x-axis and centre at origin is
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
\( \implies \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \)
\( \implies \frac{y}{x} \frac{dy}{dx} = \frac{b^2}{a^2} \)
Again differentiating both sides w.r.t. \( x \), we have
\( \frac{y}{x} \cdot \frac{d^2 y}{dx^2} + \frac{dy}{dx} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = 0 \)
\( \implies \frac{y}{x} \frac{d^2 y}{dx^2} + \frac{1}{x} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x^2} \frac{dy}{dx} = 0 \)
\( \implies xy \frac{d^2 y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \left( \frac{dy}{dx} \right) = 0 \)
which is the required differential equation.

Question. Write the solution of the differential equation \( (e^x + e^{-x}) dy = (e^x - e^{-x}) dx \)
Answer: Sol. We have, \( dy = \frac{e^x - e^{-x}}{e^x + e^{-x}} dx \)
Integrating both sides, we get
\( y = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx = \log | e^x + e^{-x} | + C \)

Question. Find the general solution of the following differential equation: \( e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0 \) 
Answer: Sol. Given differential equation,
\( e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0 \)
\( \implies e^x \tan y \, dx = - (1 - e^x) \sec^2 y \, dy \)
\( \therefore dy = \frac{e^x}{e^x - 1} \cdot \frac{\tan y}{\sec^2 y} dx \)
\( \implies \frac{\sec^2 y \, dy}{\tan y} = \frac{e^x}{e^x - 1} dx \)
Integrating both sides, we get
\( \int \frac{\sec^2 y \, dy}{\tan y} = \int \frac{e^x}{e^x - 1} dx \)
\( \implies \log | \tan y | = \log | e^x - 1 | + \log C \)
\( \implies \log | \tan y | = \log | (e^x - 1) C | \)
\( \therefore \tan y = (e^x - 1) C \)

Question. For the differential equation \( xy \frac{dy}{dx} = (x + 2) (y + 2) \), find the solution curve passing through the point (1, -1).
Answer: Sol. The given equation is \( xy \frac{dy}{dx} = (x + 2) (y + 2) \)
\( \implies \frac{y}{y + 2} dy = \frac{x + 2}{x} dx \)
\( \implies \int \frac{y}{y + 2} dy = \int \left( \frac{x + 2}{x} \right) dx \)
\( \implies \int \left( 1 - \frac{2}{y + 2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx \)
\( \implies y - 2 \log | y + 2 | = x + 2 \log | x | + C \)
\( \implies y = x + 2 \log | x | + 2 \log | y + 2 | + C \)
\( \implies y = x + 2 \log | x (y + 2) | + C \)
Since the line passes through the point (1, -1). So, putting x =1, y = -1.
We have, \( -1 = 1 + 2 \log | 1 (-1 + 2) | + C \)
\( \implies C = -2 \)
\( \therefore y = x + 2 \log | x (y + 2) | - 2 \), which is the required equation of the curve.

Question. Show that the differential equation \( 2y e^{x/y} dx + (y - 2x e^{x/y}) dy = 0 \) is homogeneous and find its particular solution, given that \( x = 0 \) when \( y = 1 \). 
Answer: Sol. Given: \( 2y \cdot e^{x/y} dx + (y - 2x e^{x/y}) dy = 0 \)
\( \implies \frac{dx}{dy} = - \frac{y - 2x e^{x/y}}{2y \cdot e^{x/y}} \)
\( \implies \frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y \cdot e^{x/y}} \)
Let \( F(x, y) = \frac{2x \cdot e^{x/y} - y}{2y \cdot e^{x/y}} \)
\( \therefore F(\lambda x, \lambda y) = \frac{2\lambda x \cdot e^{\lambda x/\lambda y} - \lambda y}{2\lambda y \cdot e^{\lambda x/\lambda y}} = \lambda^0 \frac{2x \cdot e^{x/y} - y}{2y \cdot e^{x/y}} = \lambda^0 \cdot F(x, y) \)
Hence, given differential equation is homogeneous.
Now, \( \frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y \cdot e^{x/y}} \quad \dots \text{(i)} \)
Let \( x = vy \)
\( \implies \frac{dx}{dy} = v + y \frac{dv}{dy} \)
\( \therefore \text{(i)} \implies v + y \cdot \frac{dv}{dy} = \frac{2vy \cdot e^{\frac{vy}{y}} - y}{2y \cdot e^{\frac{vy}{y}}} \)
\( \implies y \cdot \frac{dv}{dy} = \frac{y (2v e^v - 1)}{2y \cdot e^v} - v \)
\( \implies y \cdot \frac{dv}{dy} = \frac{2v \cdot e^v - 1}{2e^v} - v \)
\( \implies y \cdot \frac{dv}{dy} = - \frac{1}{2e^v} \)
\( \implies 2y e^v dv = -dy \)
\( \implies 2 \int e^v dv = - \int \frac{dy}{y} \)
\( \implies 2e^v = - \log y + C \)
\( \implies 2e^{\frac{x}{y}} + \log y = C \)
When \( x = 0, \ y = 1 \)
\( \therefore 2e^0 + \log 1 = C \text{ or } C = 2 \)
Hence, the required solution is
\( 2e^{x/y} + \log y = 2 \)
\( \implies \log C = 2 \)

Question. Show that the given differential equation is homogeneous and solve it.
\( \left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y dx = \left\{ y \sin \left( \frac{y}{x} \right) - x \cos \left( \frac{y}{x} \right) \right\} x dy \).
Answer: Sol. The given differential equation can be expressed as
\( \frac{dy}{dx} = \frac{\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y}{\left\{ y \sin \left( \frac{y}{x} \right) - x \cos \left( \frac{y}{x} \right) \right\} x} = \frac{\left\{ \cos \left( \frac{y}{x} \right) + \frac{y}{x} \sin \left( \frac{y}{x} \right) \right\} \frac{y}{x}}{\left\{ \frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right) \right\}} = f(x, y) \text{ (say)} \)
Now \( f(\lambda x, \lambda y) = \frac{\left\{ \cos \left( \frac{\lambda y}{\lambda x} \right) + \left( \frac{\lambda y}{\lambda x} \right) \sin \left( \frac{\lambda y}{\lambda x} \right) \right\} \left( \frac{\lambda y}{\lambda x} \right)}{\left\{ \left( \frac{\lambda y}{\lambda x} \right) \sin \left( \frac{\lambda y}{\lambda x} \right) - \cos \left( \frac{\lambda y}{\lambda x} \right) \right\}} = \lambda^0 f(x, y) \)
Therefore, \( f(x, y) \) is a homogeneous function of degree zero. So, the given differential equation is a homogeneous differential equation.
Put \( y = vx \) so that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
Putting values of \( \frac{dy}{dx} \) and \( y \) in the given equation, we have
\( v + x \frac{dv}{dx} = \frac{\{ x \cos v + vx \sin v \} vx}{\{ vx \sin v - x \cos v \} x} \)
\( \implies x \frac{dv}{dx} = \frac{(\cos v + v \sin v) v}{v \sin v - \cos v} - v \)
\( \implies x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} \)
\( \implies x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v} \)
\( \implies \frac{v \sin v - \cos v}{v \cos v} dv = \frac{2}{x} dx \)
\( \implies \left( \tan v - \frac{1}{v} \right) dv = \frac{2}{x} dx \)
\( \implies \int \left( \tan v - \frac{1}{v} \right) dv = 2 \int \frac{1}{x} dx \)
\( \implies \log | \sec v | - \log | v | = 2 \log x + \log C \)
\( \therefore \log | \sec v | - \log | v | - \log x^2 = \log C \)
\( \implies \log \frac{\sec v}{v x^2} = \log C \)
\( \implies \frac{\sec v}{v x^2} = C \)
\( \implies \frac{\sec \left( \frac{y}{x} \right)}{\frac{y}{x} \cdot x^2} = C \)
\( \implies \sec \left( \frac{y}{x} \right) = C x y \)
which is the required solution.

Question. Solve: \( x \frac{dy}{dx} + y - x + xy \cot x = 0 (x \neq 0) \) 
Answer: Sol. The given differential equation \( x \frac{dy}{dx} + y - x + xy \cot x = 0 (x \neq 0) \)
\( \implies \frac{dy}{dx} + \left( \cot x + \frac{1}{x} \right) y = 1 \) (Dividing both sides by x) ...(i)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where \( P = \cot x + \frac{1}{x} \) and \( Q = 1 \).
So, \( \text{IF} = e^{\int \left( \cot x + \frac{1}{x} \right) dx} = e^{\log |\sin x| + \log |x|} \)
\( = e^{\log |x \sin x|} = |x \sin x| = x \sin x \) (x sin x is always +ve for any x)
Multiplying both sides by IF in equation (i), we get
\( x \sin x \frac{dy}{dx} + x \sin x \left( \cot x + \frac{1}{x} \right) y = x \sin x \)
\( \implies x \sin x \frac{dy}{dx} + (x \cos x + \sin x) y = x \sin x \)
\( \implies \frac{d}{dx} (y x \sin x) = x \sin x \) [By product rule]
On integrating both sides, we get
\( y(x \sin x) = \int x \sin x \, dx + C \) ...(ii)
Let \( I = \int x \sin x \, dx = x \times (-\cos x) - \int 1 \cdot (-\cos x) dx \) (Using by parts)
\( I = -x \cos x + \sin x \)
Putting the value of I in (ii), we get
\( y(x \sin x) = -x \cos x + \sin x + C \)
\( \implies y(x \sin x) = \sin x - x \cos x + C \)
Hence, \( y = \frac{1}{x} - \cot x + \frac{C}{x \sin x} \) is the required solution.

Question. Find the particular solution of the differential equation \( (1 + e^{2x}) dy + (1 + y^2) e^x dx = 0 \) given that \( y = 1 \) when \( x = 0 \). 
Answer: Sol. We have, \( (1 + e^{2x}) dy + (1 + y^2) e^x dx = 0 \) and given that \( y = 1 \), when \( x = 0 \)
\( \therefore \frac{dy}{dx} = \frac{-(1 + y^2) e^x}{1 + e^{2x}} \)
\( \implies \frac{dy}{-(1 + y^2)} = \frac{e^x dx}{1 + e^{2x}} \)
Integrating both sides, we get
\( -\int \frac{dy}{1 + y^2} = \int \frac{e^x dx}{1 + e^{2x}} \)
\( \implies -\tan^{-1} y = \int \frac{e^x dx}{1 + (e^x)^2} \)
\( \implies -\tan^{-1} y = \int \frac{dt}{1 + t^2} \) [Putting \( e^x = t \)
\( \implies \) \( e^x dx = dt \)]
\( \implies -\tan^{-1} y = \tan^{-1}(t) + C \)
\( \implies -\tan^{-1} y = \tan^{-1}(e^x) + C \) ...(i)
Put x = 0, y = 1 in (i), we get
\( -\tan^{-1} 1 = \tan^{-1}(e^0) + C \)
\( \implies -\frac{\pi}{4} = \frac{\pi}{4} + C \)
\( \implies C = -\frac{\pi}{2} \)
Putting the value of C in (i), we get
\( -\tan^{-1} y = \tan^{-1}(e^x) - \frac{\pi}{2} \)
\( \implies \frac{\pi}{2} = \tan^{-1}(e^x) + \tan^{-1} y \)
Hence, \( \tan^{-1}(e^x) + \tan^{-1} y = \frac{\pi}{2} \) is the required solution.

Question. Solve that the differential equation is homogeneous and solve it. \( \left( 1 + e^{\frac{x}{y}} \right) dx + e^{\frac{x}{y}} \left( 1 - \frac{x}{y} \right) dy = 0 \)
Answer: Sol. We have, \( (1 + e^{x/y}) dx + e^{x/y} \left( 1 - \frac{x}{y} \right) dy = 0 \)
\( \implies (1 + e^{x/y}) dx = -e^{x/y} \left( 1 - \frac{x}{y} \right) dy \)
\( \therefore \frac{dx}{dy} = \frac{-e^{x/y} \left( 1 - \frac{x}{y} \right)}{(1 + e^{x/y})} = g\left(\frac{x}{y}\right) \) ...(i)
Here, RHS of differential equation is of the form \( g\left(\frac{x}{y}\right) \), so it is a homogeneous function of degree zero.
Now, we put \( x = vy \) and \( \frac{dx}{dy} = v + y \frac{dv}{dy} \)
From (i), we get \( v + y \frac{dv}{dy} = \frac{-e^v(1 - v)}{1 + e^v} \)
\( \implies y \frac{dv}{dy} = \frac{-e^v(1 - v)}{1 + e^v} - v = \frac{-(v + e^v)}{1 + e^v} \)
\( \implies \frac{1 + e^v}{-(v + e^v)} dv = \frac{dy}{y} \)
On integrating both sides, we get
\( -\log |v + e^v| + \log C = \log |y| \)
\( \implies \log C = \log |y| + \log |v + e^v| \)
\( \implies \log C = \log |y(v + e^v)| = \log \left| y \left( \frac{x}{y} + e^{x/y} \right) \right| \)
\( \implies C = y \left( \frac{x}{y} + e^{x/y} \right) \text{ or } C = x + y e^{x/y} \)
Hence, \( x + y e^{x/y} = C \) is the required solution.

Question. Solve the following differential equation: \( \frac{dy}{dx} + 2y \tan x = \sin x \), given that \( y = 0 \), when \( x = \frac{\pi}{3} \) 
Answer: Sol. Given differential equation is \( \frac{dy}{dx} + 2\tan x \cdot y = \sin x \).
Comparing it with \( \frac{dy}{dx} + Py = Q \), we get \( P = 2\tan x, Q = \sin x \)
\( \therefore \text{IF} = e^{\int 2\tan x dx} = e^{2\log \sec x} = e^{\log \sec^2 x} = \sec^2 x \quad [ \because e^{\log z} = z ] \)
Hence, general solution is \( y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx + C \).
\( y \cdot \sec^2 x = \int \sec x \cdot \tan x \, dx + C \)
\( \implies y \cdot \sec^2 x = \sec x + C \)
\( \implies y = \cos x + C \cos^2 x \)
Putting y = 0 and \( x = \frac{\pi}{3} \), we get \( 0 = \cos \frac{\pi}{3} + C \cdot \cos^2 \frac{\pi}{3} \)
\( \implies 0 = \frac{1}{2} + \frac{C}{4} \)
\( \implies C = -2 \)
\( \therefore \) Required solution is \( y = \cos x - 2 \cos^2 x \).

Question. Show that the general solution of the differential equation \( \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 \) is given by \( (x + y + 1) = A(1 - x - y - 2xy) \), where A is a parameter.
Answer: Sol. The given equation is \( \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 \).
\( \therefore \frac{dy}{y^2 + y + 1} = \frac{-dx}{x^2 + x + 1} \)
\( \implies \int \frac{dy}{y^2 + y + 1} = -\int \frac{dx}{x^2 + x + 1} \)
\( \implies \int \frac{dy}{y^2 + y + \frac{1}{4} + \left( 1 - \frac{1}{4} \right)} = -\int \frac{dx}{x^2 + x + \frac{1}{4} + \left( 1 - \frac{1}{4} \right)} \)
\( \implies \int \frac{dy}{\left( y + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = -\int \frac{dx}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \)
\( \therefore \frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{y + 1/2}{\sqrt{3}/2} \right) = -\frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{x + 1/2}{\sqrt{3}/2} \right) + C \)
\( \implies \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y + 1}{\sqrt{3}} \right) + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) = C \)
\( \implies \frac{2}{\sqrt{3}} \left[ \tan^{-1} \left( \frac{2y + 1}{\sqrt{3}} \right) + \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) \right] = C \)
\( \implies \frac{2}{\sqrt{3}} \tan^{-1} \left[ \frac{\frac{2y + 1}{\sqrt{3}} + \frac{2x + 1}{\sqrt{3}}}{1 - \left( \frac{2y + 1}{\sqrt{3}} \right) \left( \frac{2x + 1}{\sqrt{3}} \right)} \right] = C \)
\( \implies \frac{2}{\sqrt{3}} \tan^{-1} \left[ \frac{\sqrt{3} (2y + 1 + 2x + 1)}{3 - (2y + 1)(2x + 1)} \right] = C \)
\( \implies \tan^{-1} \left[ \frac{2\sqrt{3} (x + y + 1)}{2(1 - x - y - 2xy)} \right] = \frac{\sqrt{3}}{2} C \)
\( \implies \left[ \frac{2\sqrt{3} (x + y + 1)}{2(1 - x - y - 2xy)} \right] = \tan \left( \frac{\sqrt{3}}{2} C \right) \)
\( \implies \frac{x + y + 1}{1 - x - y - 2xy} = \frac{1}{\sqrt{3}} \tan \left( \frac{\sqrt{3}}{2} C \right) = A \)
\( \implies (x + y + 1) = A (1 - x - y - 2xy) \).

Question. Solve the differential equation \( y e^{x/y} dx = (x e^{x/y} + y^2) dy \ (y \neq 0) \).
Answer: Sol. The given equation is \( y e^{x/y} dx = (x e^{x/y} + y^2) dy \).
\( \therefore \frac{dx}{dy} = \frac{x e^{x/y} + y^2}{y e^{x/y}} \), is a homogeneous differential equation. ...(i)
Put \( x = vy \) so that \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).
\( \therefore v + y \frac{dv}{dy} = \frac{vy e^v + y^2}{y e^v} \)
\( \implies v + y \frac{dv}{dy} = \frac{y (v e^v + y)}{y e^v} \)
\( \implies y \frac{dv}{dy} = \frac{v e^v + y}{e^v} - v \)
\( \implies y \frac{dv}{dy} = \frac{v e^v + y - v e^v}{e^v} \)
\( \implies y \frac{dv}{dy} = \frac{y}{e^v} \)
\( \implies e^v dv = dy \)
\( \implies \int e^v dv = \int dy \)
\( \implies e^v = y + C \)
\( \implies e^{x/y} = y + C \)

Question. Find the particular solution of the differential equation: \( \frac{dy}{dx} + y \cot x = 4x \operatorname{cosec} x \ (x \neq 0) \), given that \( y = 0 \) when \( x = \frac{\pi}{2} \)
Answer: Sol. Given differential equation is \( \frac{dy}{dx} + y \cot x = 4x \operatorname{cosec} x \).
It is of the type \( \frac{dy}{dx} + Py = Q \), where \( P = \cot x \), \( Q = 4x \operatorname{cosec} x \).
\( \therefore \text{IF} = e^{\int P dx} = e^{\int \cot x dx} = e^{\log |\sin x|} = \sin x \)
Its solution is given by
\( \sin x \cdot y = \int 4x \operatorname{cosec} x \cdot \sin x dx \)
\( \implies y \sin x = \int 4x dx = \frac{4x^2}{2} + C \)
\( \implies y \sin x = 2x^2 + C \) ...(i)
Now, put \( y = 0 \) when \( x = \frac{\pi}{2} \) in (i), we get
\( \therefore 0 = 2 \times \frac{\pi^2}{4} + C \)
\( \implies C = -\frac{\pi^2}{2} \)
Hence, the particular solution of given differential equation is \( y \sin x = 2x^2 - \frac{\pi^2}{2} \)

Question.
Answer: \( (1 + x^2) \frac{dx}{dy} + y = \tan^{-1} x \)

\( \implies \) \( \frac{dy}{dx} + \frac{1}{1 + x^2} \cdot y = \frac{\tan^{-1} x}{1 + x^2} \)
\( \therefore \) IF = \( e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} \)
Solution will be \( y \times e^{\tan^{-1} x} = \int \frac{\tan^{-1} x}{1 + x^2} \times e^{\tan^{-1} x} dx \dots (i) \)
Let \( e^{\tan^{-1} x} = t \)

\( \implies \) \( \frac{e^{\tan^{-1} x}}{1 + x^2} dx = dt \) and \( \log (e^{\tan^{-1} x}) = \log t \)

\( \implies \) \( \tan^{-1} x = \log t \)
From equation (i), \( \int \log t \cdot dt = t \log t - t + C \)
\( y e^{\tan^{-1} x} = e^{\tan^{-1} x}(\tan^{-1} x - 1) + C \)
Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Hence, (b) is the correct option.

Question.
Answer: Clearly, Assertion (A) is false and Reason (R) is true.
Hence, (d) is the correct option

Question.
Answer: \( \frac{dx}{dy} + (\tan y) \cdot x = \sec^2 y \)
Here, IF = \( e^{\int \tan y dy} = e^{\log \sec y} = \sec y \)
Clearly, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Hence, (a) is the correct option.

Very Short Answer Questions 

Question. What is the degree of the following differential equation
\( 5x \left( \frac{dy}{dx} \right)^2 - \frac{d^2y}{dx^2} - 6y = \log x \)?
Answer: Degree of differential equation is 1 because power of highest order derivative \( \frac{d^2y}{dx^2} \) is one.

Question. Find the order and degree of differential equation:
\( \frac{d^4y}{dx^4} + \sin \left( \frac{d^3y}{dx^3} \right) = 0 \) 
Answer: Order is 4 but degree is not defined because given differential equation cannot be written in the form of polynomial in differential co-efficient.

Question. Find the differential equation representing the curve \( y = cx + c^2 \). 
Answer: Given \( y = cx + c^2 \)
\( \implies \) \( \frac{dy}{dx} = c + 0 \)
\( \implies \) \( \frac{dy}{dx} = c \) [Differentiating with respect to x]
Again differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = 0 \)

Question. Find the differential equation representing the curve \( y = e^{-x} + ax + b \), where a and b are arbitrary constants. 
Answer: Given curve is \( y = e^{-x} + ax + b \).
\( \implies \) \( \frac{dy}{dx} = -e^{-x} + a \) [Differentiating with respect to x]
\( \implies \) \( \frac{d^2y}{dx^2} = e^{-x} \) [Differentiating again with respect to x]

Question. Find the differential equation representing the family of curves \( v = \frac{A}{r} + B \), where A and B are arbitrary constants. 
Answer: Given family of curve is \( v = \frac{A}{r} + B \).
\( \frac{dv}{dr} = \frac{-A}{r^2} \) [Differentiating with respect to r]
\( \frac{d^2v}{dr^2} = \frac{2A}{r^3} \)
\( \implies \) \( \frac{d^2v}{dr^2} = \frac{2}{r} \cdot \frac{A}{r^2} \) [Again differentiating with respect to r]
\( \implies \) \( \frac{d^2v}{dr^2} = \frac{2}{r} \left( -\frac{dv}{dr} \right) \)
\( \implies \) \( \frac{d^2v}{dr^2} = - \frac{2}{r} \frac{dv}{dr} \)
\( \implies \) \( r \frac{d^2v}{dr^2} + 2 \frac{dv}{dr} = 0 \)

Question. Write the sum of the order and degree of the following differential equation:
\( \frac{d}{dx} \left\{ \left( \frac{dy}{dx} \right)^3 \right\} = 0 \)
Answer: Given differential equation is
\( \frac{d}{dx} \left[ \left( \frac{dy}{dx} \right)^3 \right] = 0 \)

\( \implies \) \( 3 \left( \frac{dy}{dx} \right)^2 \frac{d^2y}{dx^2} = 0 \)
i.e., order = 2, degree = 1
\( \therefore \) Required sum = 2 + 1 = 3.

Short Answer Questions

Question. Write the integrating factor of the following differential equation:
\( (1 + y^2) + (2xy - \cot y) \frac{dy}{dx} = 0 \) 
Answer: \( (1 + y^2) + (2xy - \cot y) \frac{dy}{dx} = 0 \)
\( \implies \) \( (2xy - \cot y) \frac{dy}{dx} = -(1 + y^2) \)
\( \implies \) \( \frac{dy}{dx} = -\frac{1 + y^2}{2xy - \cot y} \)
\( \implies \) \( \frac{dx}{dy} = -\frac{(2xy - \cot y)}{1 + y^2} \)
\( \implies \) \( \frac{dx}{dy} + \frac{2y}{1 + y^2} \cdot x = \frac{\cot y}{1 + y^2} \)
It is in the form \( \frac{dx}{dy} + Px = Q \), where P and Q are function of y.
\( \implies \) IF = \( e^{\int P dy} = e^{\int \frac{2y}{1 + y^2} dy} = e^{\log |1 + y^2|} = 1 + y^2 \)

Question. Write the general solution of the differential equation \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: We have, \( \frac{dy}{dx} = \frac{y}{x} \)
\( \implies \) \( \frac{dy}{y} = \frac{dx}{x} \)
Integrating both sides, we get
\( \log |y| = \log |x| + \log |C| \)
\( \implies \) \( |y| = |xC| \)
\( \implies \) \( y = Cx \)

Question. Find the differential equation of the family of curves represented by \( y^2 = a(b^2 - x^2) \).
Answer: We have, \( y^2 = a(b^2 - x^2) = ab^2 - ax^2 \)
Differentiating with respect to x, we get
\( 2y \frac{dy}{dx} = -2ax \)
\( \implies \) \( y \frac{dy}{dx} = -ax \dots (i) \)
\( \implies \) \( \frac{y}{x} \frac{dy}{dx} = -a \dots (ii) \)
Again differentiating (i) with respect to x, we get
\( y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = -a \)
Using (ii), we get
\( y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = \frac{y}{x} \frac{dy}{dx} \)
\( \implies \) \( xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \)

Question. Solve the differential equation \( (y + 3x^2) \frac{dx}{dy} = x \).
Answer: \( (y + 3x^2)dx = x dy \)
\( \implies \) \( y dx + 3x^2 dx = x dy \)
\( \implies \) \( 3x^2 dx = x dy - y dx \)
\( \implies \) \( 3dx = \frac{x dy - y dx}{x^2} = d\left(\frac{y}{x}\right) \)
Integrating, we get
\( \implies \) \( 3x = \frac{y}{x} + C \)
\( \implies \) \( 3x^2 = y + Cx \)
\( \implies \) \( y - 3x^2 + Cx = 0 \).

Question. For a differential equation representing the family of curves \( y = A \sin x \), by eliminating the arbitrary constant.
Answer: We have \( y = A \sin x \dots (i) \)
Differentiating with respect to x, we get
\( \frac{dy}{dx} = A \cos x \)
Again differentiating with respect to x, we get
\( \frac{d^2y}{dx^2} = -A \sin x = -y \dots (\text{using } (i)) \)
\( \implies \) \( \frac{d^2y}{dx^2} + y = 0 \)

Question. Find the general solution of \( y^2 dx + (x^2 - xy + y^2) dy = 0 \). [NCERT Exemplar]
Answer: Given, differential equation is \( y^2 dx + (x^2 - xy + y^2) dy = 0 \).
\( \implies \) \( y^2 dx = -(x^2 - xy + y^2) dy \)
\( \implies \) \( y^2 \frac{dx}{dy} = -(x^2 - xy + y^2) \)
\( \implies \) \( \frac{dx}{dy} = -\left( \frac{x^2}{y^2} - \frac{x}{y} + 1 \right) \dots (i) \)
Which is a homogeneous differential equation.
Put \( \frac{x}{y} = v \) or \( x = vy \)
\( \implies \) \( \frac{dx}{dy} = v + y \frac{dv}{dy} \)
On substituting these values in equation (i), we get
\( v + y \frac{dv}{dy} = -[v^2 - v + 1] \)
\( \implies \) \( y \frac{dv}{dy} = -v^2 + v - 1 - v \)
\( \implies \) \( y \frac{dv}{dy} = -v^2 - 1 \)
\( \implies \) \( \frac{dv}{v^2 + 1} = -\frac{dy}{y} \)
On integrating both sides, we get
\( \tan^{-1} (v) = -\log y + C \)
\( \implies \) \( \tan^{-1} \left( \frac{x}{y} \right) + \log y = C \) \( \left[ \because v = \frac{x}{y} \right] \)