CBSE Class 12 Mathematics Differential Equations Previous Year Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Find the sum of the order and the degree of the following differential equation: \( y = x\left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} \) [CBSE 2015, 1M]
Answer:
Order = 2 and degree = 1, and sum = 2 + 1 = 3. [1]

Question 2. Find the solution of the following differential equation: \( x\sqrt{1 + y^2}\,dx + y\sqrt{1 + x^2}\,dy = 0 \) [CBSE 2015, 1M]
Answer:
\( \int \frac{x}{\sqrt{1 + x^2}}\,dx = -\int \frac{y\,dy}{\sqrt{1 + y^2}} \)
⇒ \( \frac{1}{2}\int \frac{2x}{\sqrt{1 + x^2}}\,dx = -\frac{1}{2}\int \frac{2y}{\sqrt{1 + y^2}}\,dy \)
⇒ \( \frac{1}{2} \times 2\sqrt{1 + x^2} = -\frac{1}{2} \times 2\sqrt{1 + y^2} + C \)
⇒ \( \sqrt{1 + x^2} + \sqrt{1 + y^2} = C \) [1]

Question 3. Find the general solution of the differential equation \( e^{y - x}\frac{dy}{dx} = 1 \). [CBSE 2020, 1M]
Answer:
Given \( e^{y - x}\frac{dy}{dx} = 1 \)
\( e^y \cdot e^{-x}\frac{dy}{dx} = 1 \)
\( e^y\,dy = e^x\,dx \)
On integrating both sides, we have:
\( \int e^y\,dy = \int e^x\,dx \)
\( e^y = e^x + C \) [1]

Long Answer Questions I (4 Marks)

Question 4. Solve the differential equation \( (1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1} x} \). [CBSE 2014, 4M]
Answer:
\( (1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1} x} \)
\( \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{1 + x^2} \)
Let \( P = \frac{1}{1 + x^2} \) and \( Q = \frac{e^{\tan^{-1} x}}{1 + x^2} \) [½]
\( \text{IF} = e^{\int P\,dx} = e^{\int \frac{1}{1 + x^2}dx} = e^{\tan^{-1} x} \) [1]
Solution of the D.E.:
\( y \times e^{\tan^{-1} x} = \int \frac{e^{\tan^{-1} x}}{1 + x^2} \times e^{\tan^{-1} x}\,dx \) [1]
⇒ \( y\,e^{\tan^{-1} x} = \int \frac{(e^{\tan^{-1} x})^2}{1 + x^2}\,dx \)   [Put \( \tan^{-1} x = t \Rightarrow \frac{1}{1 + x^2}dx = dt \)] [½]
⇒ \( y\,e^{\tan^{-1} x} = \int e^{2t}\,dt = \frac{e^{2t}}{2} + C \)
⇒ \( y\,e^{\tan^{-1} x} = \frac{e^{2\tan^{-1} x}}{2} + C \) [1]

Question 5. Find the particular solution of the differential equation \( \log\left(\frac{dy}{dx}\right) = 3x + 4y \), given that y = 0 when x = 0. [CBSE 2014, 4M]
Answer:
\( \log\left(\frac{dy}{dx}\right) = 3x + 4y \)
⇒ \( \frac{dy}{dx} = e^{3x + 4y} \)
⇒ \( \frac{dy}{dx} = e^{3x} \cdot e^{4y} \)
⇒ \( \int e^{-4y}\,dy = \int e^{3x}\,dx \)
⇒ \( \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C \) ...(i) [2]
Given that y = 0 when x = 0,
\( C = \frac{-1}{4} - \frac{1}{3} = \frac{-7}{12} \) [1]
⇒ \( \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12} \) (from equation (i))
or \( 3e^{-4y} + 4e^{3x} = 7 \) [1]

Question 6. Find the particular solution of the differential equation \( 2ye^{x/y}\,dx + (y - 2xe^{x/y})\,dy = 0 \), given that x = 0 when y = 1. [CBSE 2016, 4M]
Answer:
\( 2ye^{x/y}\,dx + (y - 2xe^{x/y})\,dy = 0 \)
\( \frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}} \)
Putting \( \frac{x}{y} = v \), i.e. \( x = vy \Rightarrow \frac{dx}{dy} = v + y\frac{dv}{dy} \) [1]
⇒ \( v + y\frac{dv}{dy} = \frac{2vye^v - y}{2ye^v} \)
or \( v + y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} \)
⇒ \( y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} - v \)
or \( y\frac{dv}{dy} = \frac{2ve^v - 1 - 2ve^v}{2e^v} \)
⇒ \( y\frac{dv}{dy} = \frac{-1}{2e^v} \)
or \( 2\int e^v\,dv = -\int \frac{dy}{y} \) [1]
⇒ \( 2e^v = -\log y + C \)
⇒ \( 2e^{x/y} = -\log y + C \) ...(1) [1]
When x = 0, y = 1:
⇒ 2 = 0 + C ⇒ C = 2
∴ \( 2e^{x/y} + \log y = 2 \) [1]

Question 7. Find the particular solution of the differential equation \( \frac{dy}{dx} = -\frac{x + y\cos x}{1 + \sin x} \), given that y = 1 when x = 0. [CBSE 2016, 4M]
Answer:
\( \frac{dy}{dx} = -\frac{x + y\cos x}{1 + \sin x} \)
\( \frac{dy}{dx} + \left(\frac{\cos x}{1 + \sin x}\right)y = -\frac{x}{1 + \sin x} \) ...(i) [1]
Eq. (i) is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \),
where \( P = \frac{\cos x}{1 + \sin x} \) and \( Q = -\frac{x}{1 + \sin x} \)
⇒ \( \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{\cos x}{1 + \sin x}dx} = e^{\log(1 + \sin x)} = 1 + \sin x \) [1]
Hence, the solution of diff. eq. (i) is
\( y \times \text{I.F.} = \int (Q \times \text{I.F.})\,dx \)
⇒ \( y(1 + \sin x) = \int -\frac{x}{1 + \sin x} \times (1 + \sin x)\,dx \)
⇒ \( y(1 + \sin x) = -\frac{x^2}{2} + C \) [1]
When x = 0, y = 1 ⇒ C = 1
∴ \( y(1 + \sin x) = -\frac{x^2}{2} + 1 \) [1]

Question 8. 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. [CBSE 2018, 4M]
OR
Find the particular solution of the differential equation \( \frac{dy}{dx} + 2y\tan x = \sin x \), given that y = 0 when \( x = \frac{\pi}{3} \).
Answer:
The given differential equation is
\( e^x\tan y\,dx + (2 - e^x)\sec^2 y\,dy = 0 \)
∴ \( (2 - e^x)\sec^2 y\,dy = -e^x\tan y\,dx \)
⇒ \( \frac{\sec^2 y}{\tan y}\,dy = \frac{-e^x}{2 - e^x}\,dx \) [1]
Integrating both sides, we get:
\( \int \frac{\sec^2 y}{\tan y}\,dy = \int \frac{-e^x}{2 - e^x}\,dx \) ...(1)
⇒ \( \log(\tan y) = \log(2 - e^x) + \log C \)
⇒ \( \log(\tan y) = \log[C(2 - e^x)] \)
⇒ \( \tan y = C(2 - e^x) \) [1]
When \( y = \frac{\pi}{4} \), x = 0 ⇒ C = 1
So, the required particular solution is \( \tan y = 2 - e^x \).
OR
The given differential equation is \( \frac{dy}{dx} + 2y\tan x = \sin x \)
This is a linear equation of the form \( \frac{dy}{dx} + Py = Q \),
where P = 2 tan x and Q = sin x.
Now, \( \text{I.F.} = e^{\int P\,dx} = e^{\int 2\tan x\,dx} = e^{2\log|\sec x|} = e^{\log|\sec^2 x|} = \sec^2 x \) [1]
The general solution of the given differential equation is
\( y(\text{I.F.}) = \int (Q \times \text{I.F.})\,dx + C \)
⇒ \( y(\sec^2 x) = \int (\sin x \cdot \sec^2 x)\,dx + C \) [1]
⇒ \( y\sec^2 x = \int (\sec x \cdot \tan x)\,dx + C \)
⇒ \( y\sec^2 x = \sec x + C \) ...(1)
Now, y = 0, \( x = \frac{\pi}{3} \).
Therefore, \( 0 \times \sec^2\frac{\pi}{3} = \sec\frac{\pi}{3} + C \) [1]
⇒ 0 = 2 + C
⇒ C = −2
Substituting C = −2 in equation (1), we get:
\( y\sec^2 x = \sec x - 2 \)
⇒ \( y = \cos x - 2\cos^2 x \) [1]
Hence, the required solution of the given differential equation is \( y = \cos x - 2\cos^2 x \).

Question 9. Solve the differential equation: \( x\frac{dy}{dx} = y - x\tan\left(\frac{y}{x}\right) \) [CBSE 2019, 4M]
OR
Solve the differential equation: \( \frac{dy}{dx} = -\left[\frac{x + y\cos x}{1 + \sin x}\right] \)
Answer:
\( x\frac{dy}{dx} = y - x\tan(y/x) \)
⇒ \( \frac{dy}{dx} = \left(\frac{y}{x}\right) - \tan\left(\frac{y}{x}\right) \) ...(i) [½]
The given equation is a homogeneous differential equation.
Put \( \frac{y}{x} = v \), i.e. \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \) [1]
∴ \( v + x\frac{dv}{dx} = v - \tan v \)
⇒ \( x\frac{dv}{dx} = -\tan v \)
⇒ \( \frac{dv}{\tan v} = -\frac{dx}{x} \) [1]
Integrating both sides, we get:
\( \int \frac{dv}{\tan v} = \int \frac{-dx}{x} \)
⇒ \( \int \cot v\,dv = -\int \frac{dx}{x} \)
⇒ log |sin v| = −log |x| + log C
⇒ \( \log|\sin v| = \log\left|\frac{C}{x}\right| \) [1]
⇒ \( \sin\left(\frac{y}{x}\right) = \frac{C}{x} \)
or \( x\sin\left(\frac{y}{x}\right) = C \) [½]
OR
\( \frac{dy}{dx} = -\left[\frac{x + y\cos x}{1 + \sin x}\right] \)
⇒ \( \frac{dy}{dx} + \left(\frac{\cos x}{1 + \sin x}\right)y = \frac{-x}{1 + \sin x} \) ...(i) [1½]
Eq. (i) is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \),
where \( P = \left(\frac{\cos x}{1 + \sin x}\right) \) and \( Q = \left(\frac{-x}{1 + \sin x}\right) \)
⇒ \( \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{\cos x}{1 + \sin x}dx} = e^{\log|1 + \sin x|} = 1 + \sin x \) [1]
Hence, the solution is
\( y \times \text{I.F.} = \int Q \times \text{I.F.}\,dx + C \)
⇒ \( y \times (1 + \sin x) = \int \left(\frac{-x}{1 + \sin x}\right) \times (1 + \sin x)\,dx + C \) [1]
⇒ \( y(1 + \sin x) = \int -x\,dx + C \)
or \( y(1 + \sin x) = \frac{-x^2}{2} + C \) [½]

Question 10. Solve the following differential equation: \( (1 + e^{y/x})\,dy + e^{y/x}\left(1 - \frac{y}{x}\right)dx = 0\ (x \neq 0) \). [CBSE 2020, 4M]
Answer:
Given \( (1 + e^{y/x})\,dy + e^{y/x}\left(1 - \frac{y}{x}\right)dx = 0 \)
⇒ \( (1 + e^{y/x})\,dy = \left(\frac{y}{x} - 1\right)e^{y/x}\,dx \)
⇒ \( \frac{dy}{dx} = \frac{e^{y/x}\left(\frac{y}{x} - 1\right)}{1 + e^{y/x}} \) [1]
Put \( \frac{y}{x} = v \), i.e. \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \)
\( v + x\frac{dv}{dx} = \frac{e^v(v - 1)}{1 + e^v} \Rightarrow x\frac{dv}{dx} = \frac{e^v(v - 1)}{1 + e^v} - v \) [1]
⇒ \( x\frac{dv}{dx} = \frac{ve^v - e^v - v - ve^v}{1 + e^v} \)
⇒ \( x\frac{dv}{dx} = -\frac{(e^v + v)}{1 + e^v} \)
⇒ \( \int \frac{1 + e^v}{v + e^v}\,dv = -\int \frac{dx}{x} \) [1]
⇒ \( \log|v + e^v| = -\log|x| + \log C \)
⇒ \( \log|v + e^v| = \log\left|\frac{C}{x}\right| \)
⇒ \( v + e^v = \frac{C}{x} \)
⇒ \( x\left(\frac{y}{x} + e^{y/x}\right) = C \)
⇒ \( y + xe^{y/x} = C \) [1]

Long Answer Questions II (6 Marks)

Question 11. Find the particular solution of the differential equation \( (x - y)\frac{dy}{dx} = (x + 2y) \), given that y = 0 when x = 1. [CBSE 2017, 6M]
Answer:
The given differential equation can be expressed as:
\( \frac{dy}{dx} = \frac{x + 2y}{x - y} \) ...(1) [½]
Let \( F(x, y) = \frac{x + 2y}{x - y} \)
Now \( F(\lambda x, \lambda y) = \frac{\lambda(x + 2y)}{\lambda(x - y)} = \lambda^0 \cdot F(x, y) \)
Therefore, F(x, y) is a homogeneous function of degree zero. So, the given differential equation is a homogeneous differential equation.
Alternatively,
\( \frac{dy}{dx} = \left(\frac{1 + \frac{2y}{x}}{1 - \frac{y}{x}}\right) = g\left(\frac{y}{x}\right) \) ...(2)
R.H.S. of differential equation (2) is of the form \( g\left(\frac{y}{x}\right) \) and so it is a homogeneous function of degree zero. Therefore, equation (1) is a homogeneous differential equation.
Put \( \frac{y}{x} = v \), i.e. \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \) [½]
\( v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} \) [½]
or \( x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v \)
or \( x\frac{dv}{dx} = \frac{v^2 + v + 1}{1 - v} \)
or \( \frac{v - 1}{v^2 + v + 1}\,dv = \frac{-dx}{x} \) [1]
Integrating both sides, we get
\( \int \frac{v - 1}{v^2 + v + 1}\,dv = -\int \frac{dx}{x} \)
or \( \frac{1}{2}\int \frac{2v + 1 - 3}{v^2 + v + 1}\,dv = -\int \frac{1}{x}\,dx \)
or \( \frac{1}{2}\int \frac{2v + 1}{v^2 + v + 1}\,dv - \frac{3}{2}\int \frac{1}{v^2 + v + 1}\,dv = -\int \frac{1}{x}\,dx \)
or \( \frac{1}{2}\int \frac{2v + 1}{v^2 + v + 1}\,dv - \frac{3}{2}\int \frac{1}{\left(v + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\,dv = -\int \frac{1}{x}\,dx \) [1]
or \( \frac{1}{2}\log|v^2 + v + 1| - \frac{3}{2} \cdot \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = -\log|x| + C_1 \)
or \( \frac{1}{2}\log|v^2 + v + 1| + \frac{1}{2}\log x^2 = \sqrt{3}\,\tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) + C_1 \) [1]
Replacing v by \( \frac{y}{x} \), we get
\( \frac{1}{2}\log\left|\frac{y^2}{x^2} + \frac{y}{x} + 1\right| + \frac{1}{2}\log x^2 = \sqrt{3}\,\tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) + C_1 \) [½]
or \( \frac{1}{2}\log\left|\left(\frac{y^2}{x^2} + \frac{y}{x} + 1\right)x^2\right| = \sqrt{3}\,\tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) + C_1 \)
or \( \log|y^2 + xy + x^2| = 2\sqrt{3}\,\tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) + 2C_1 \)
or \( \log|y^2 + xy + x^2| = 2\sqrt{3}\,\tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) + C \) ...(3) [½]
When y = 0, x = 1, then (putting in equation (3))
\( 0 = 2\sqrt{3}\left(\frac{\pi}{6}\right) + C \Rightarrow C = -\frac{\sqrt{3}\pi}{3} = -\frac{\pi}{\sqrt{3}} \)
⇒ \( \log|y^2 + xy + x^2| = 2\sqrt{3}\,\tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) - \frac{\pi}{\sqrt{3}} \)
which is the required particular solution of the differential equation (1). [½]