CBSE Class 12 Mathematics Differential Equations Class Test Set 01

Class Test Questions with Solutions

Question 1. Find the order and degree of the differential equation \( \frac{d^4y}{dx^4} + \sin\left(\frac{d^3y}{dx^3}\right) = 0 \). [1]
Answer:
The highest order derivative present in the given differential equation is 4, so the order of the given differential equation is 4. As it is not expressible as a polynomial in differential coefficients, its degree is not defined.

Question 2. Find the order and degree of the differential equation \( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} = 2 \). [1]
Answer:
The given differential equation when written as a polynomial in derivatives becomes
\( 4\left(\frac{d^2y}{dx^2}\right)^2 = \left\{1 + \left(\frac{dy}{dx}\right)^2\right\}^3 \)
The highest order differential coefficient in this equation is \( \frac{d^2y}{dx^2} \) and its power is 2. Therefore, the given differential equation is a non-linear differential equation of second order and second degree.

Question 3. Find the integrating factor of the differential equation \( (\tan^{-1} y - x)\,dy = (1 + y^2)\,dx \). [1]
Answer:
\( \frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2} \)
⇒ \( \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2} \)
⇒ \( \frac{dx}{dy} + Px = Q \), where \( P = \frac{1}{1 + y^2} \), \( Q = \frac{\tan^{-1} y}{1 + y^2} \)
\( \text{I.F.} = e^{\int P\,dy} = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1} y} \)

Question 4. Find the solution of the differential equation \( \frac{dy}{dx} + \frac{1}{y} + y = 0 \). [2]
Answer:
\( \frac{dy}{dx} + \frac{1}{y} + y = 0 \)
⇒ \( \frac{dy}{dx} = -\left(\frac{1 + y^2}{y}\right) \)
or \( \int \frac{y\,dy}{1 + y^2} = -\int dx \)
Put \( 1 + y^2 = t \)
⇒ 2y dy = dt
∴ \( y\,dy = \frac{dt}{2} \)
⇒ \( \frac{1}{2}\int \frac{dt}{t} = -\int dx \)
or \( \frac{1}{2}\log|t| = -x + C \)
⇒ \( x + \frac{1}{2}\log|1 + y^2| = C \)

Question 5. Find the solution of the following differential equation: \( x\sqrt{1 + y^2}\,dx + y\sqrt{1 + x^2}\,dy = 0 \). [2]
Answer:
\( \int \frac{x}{\sqrt{1 + x^2}}\,dx = -\int \frac{y\,dy}{\sqrt{1 + y^2}} \)
⇒ \( \frac{1}{2}\int \frac{2x}{\sqrt{1 + x^2}}\,dx = -\frac{1}{2}\int \frac{2y}{\sqrt{1 + y^2}}\,dy \)
⇒ \( \frac{1}{2} \times 2\sqrt{1 + x^2} = -\frac{1}{2} \times 2\sqrt{1 + y^2} + C \)
⇒ \( \sqrt{1 + x^2} + \sqrt{1 + y^2} = C \)

Question 6. Solve: (sin x + cos x) dy + (cos x − sin x) dx = 0 [2]
Answer:
(sin x + cos x) dy = (sin x − cos x) dx
\( \int dy = \int \frac{(\sin x - \cos x)}{(\sin x + \cos x)}\,dx \)   (integrating both sides)
Put sin x + cos x = t
∴ (cos x − sin x) dx = dt or (sin x − cos x) dx = −dt
or \( \int dy = -\int \frac{dt}{t} \)
⇒ y = −log |t| + C = −log |(sin x + cos x)| + C
⇒ \( e^y(\sin x + \cos x) = C \)

Question 7. Solve the following differential equation: \( (x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4} \). [3]
Answer:
The given differential equation is
\( (x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4} \)
⇒ \( \frac{dy}{dx} + \left(\frac{2x}{x^2 + 1}\right)y = \frac{\sqrt{x^2 + 4}}{x^2 + 1} \)
which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here \( P = \frac{2x}{x^2 + 1} \), \( Q = \frac{\sqrt{x^2 + 4}}{x^2 + 1} \)
∴ Integrating factor, \( \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{2x}{x^2 + 1}dx} \)
Put \( x^2 + 1 = t \Rightarrow 2x\,dx = dt \)
\( \text{I.F.} = e^{\int \frac{dt}{t}} = e^{\log t} = t = x^2 + 1 \)
∴ The required solution is \( y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C \)
⇒ \( y(x^2 + 1) = \int \frac{\sqrt{x^2 + 4}}{x^2 + 1}(x^2 + 1)\,dx + C \)
∴ \( y(x^2 + 1) = \int \sqrt{x^2 + 4}\,dx + C \)   \( \left[\because \int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C\right] \)
⇒ \( y(x^2 + 1) = \frac{x}{2}\sqrt{x^2 + 4} + \frac{4}{2}\log\left|x + \sqrt{x^2 + 4}\right| + C \)
or \( y(x^2 + 1) = \frac{x}{2}\sqrt{x^2 + 4} + 2\log\left|x + \sqrt{x^2 + 4}\right| + C \)

Question 8. Solve the differential equation: \( \frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x} \). [3]
Answer:
The differential equation is of the form \( \frac{dy}{dx} + Py = Q \),
where \( P = -\frac{1}{x} \) and \( Q = \frac{\log x}{x} \)
∴ \( \text{I.F.} = e^{\int P\,dx} \)
Its solution is \( y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + C \)
or \( y\,e^{-\int \frac{dx}{x}} = \int \frac{\log x}{x}\,e^{-\int \frac{dx}{x}}\,dx + C \)
⇒ \( y\,e^{-\ln x} = \int \frac{\log x}{x} \times e^{-\ln x}\,dx + C \)
or \( \frac{y}{x} = \int \underset{\text{I}}{\log x} \cdot \underset{\text{II}}{\frac{1}{x^2}}\,dx + C \)
∴ \( \frac{y}{x} = \frac{\log x\,(x^{-1})}{(-1)} - \int \frac{1}{x} \cdot \frac{x^{-1}}{(-1)}\,dx + C \)
⇒ \( \frac{y}{x} = -\frac{\log x}{x} - \frac{1}{x} + C \), i.e. \( y = Cx - \log x - 1 \)

Question 9. Solve the following differential equation: \( \frac{dy}{dx} + 2y\tan x = \sin x \). [3]
Answer:
The given differential equation is:
\( \frac{dy}{dx} + 2y\tan x = \sin x \)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here P = 2 tan x, Q = sin x
\( \text{I.F.} = e^{\int P\,dx} = e^{\int 2\tan x\,dx} = e^{2\log\sec x} = e^{\log\sec^2 x} = \sec^2 x \)
∴ The required solution is \( y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C \)
⇒ \( y\sec^2 x = \int \sin x \cdot \sec^2 x\,dx + C \)
or \( y\sec^2 x = \int \frac{\sin x}{\cos^2 x}\,dx + C \)
⇒ \( y\sec^2 x = \int \sec x\tan x\,dx + C \)
or \( y\sec^2 x = \sec x + C \)
⇒ \( y = \frac{\sec x}{\sec^2 x} + \frac{C}{\sec^2 x} \)
∴ \( y = \cos x + C\cos^2 x \)

Question 10. Solve: \( (x^3 - 3xy^2)\,dx = (y^3 - 3x^2y)\,dy \). [5]
Answer:
We are given that
\( (x^3 - 3xy^2)\,dx = (y^3 - 3x^2y)\,dy \)
⇒ \( \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \)
Clearly, the given equation is a homogeneous equation. Putting y = vx and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) in it, we get
\( v + x\frac{dv}{dx} = \frac{x^3 - 3v^2x^3}{v^3x^3 - 3vx^3} \)
⇒ \( v + x\frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} \)
⇒ \( x\frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} - v = \frac{1 - v^4}{v^3 - 3v} \)
⇒ \( x(v^3 - 3v)\,dv = (1 - v^4)\,dx \)
⇒ \( \frac{v^3 - 3v}{1 - v^4}\,dv = \frac{dx}{x} \)
⇒ \( \int \frac{v^3 - 3v}{1 - v^4}\,dv = \int \frac{dx}{x} \)
⇒ \( \int \frac{v^3}{1 - v^4}\,dv - 3\int \frac{v}{1 - v^4}\,dv = \int \frac{dx}{x} \)
⇒ \( -\frac{1}{4}\int \frac{-4v^3}{1 - v^4}\,dv - \frac{3}{2}\int \frac{2v}{1 - (v^2)^2}\,dv = \int \frac{dx}{x} \)
⇒ \( -\frac{1}{4}\int \frac{-4v^3}{1 - v^4}\,dv - \frac{3}{2}\int \frac{dt}{1 - t^2} = \int \frac{dx}{x} \), where \( t = v^2 \)
⇒ \( -\frac{1}{4}\log|1 - v^4| - \frac{3}{2} \times \frac{1}{2 \times 1}\log\left|\frac{1 + t}{1 - t}\right| = \log|x| + \log C \)
⇒ \( -\frac{1}{4}\log|1 - v^4| - \frac{3}{4}\log\left|\frac{1 + v^2}{1 - v^2}\right| = \log|Cx| \)
⇒ \( -\log|1 - v^4| - 3\log\left|\frac{1 + v^2}{1 - v^2}\right| = 4\log|Cx| \)
⇒ \( \log\left|(1 - v^4)^{-1}\left(\frac{1 + v^2}{1 - v^2}\right)^{-3}\right| = \log|(Cx)^4| \)
⇒ \( \frac{1}{1 - v^4} \times \left(\frac{1 - v^2}{1 + v^2}\right)^3 = (Cx)^4 \)
⇒ \( (1 - v^2)^2 = (1 + v^2)^4(Cx)^4 \)
⇒ \( 1 - v^2 = (1 + v^2)^2(Cx)^2 \)
⇒ \( 1 - \frac{y^2}{x^2} = \left(1 + \frac{y^2}{x^2}\right)^2C^2x^2 \)
⇒ \( x^2 - y^2 = (x^2 + y^2)^2C^2 \), which is the required solution.

Question 11. Show that the given differential equation is homogeneous and solve it: \( \left(1 + e^{\frac{x}{y}}\right)dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0 \) [5]
Answer:
\( (1 + e^{x/y})\,dx + e^{x/y}\left(1 - \frac{x}{y}\right)dy = 0 \)
⇒ \( \frac{dy}{dx} = -\frac{1 + e^{x/y}}{e^{x/y}\left(1 - \frac{x}{y}\right)} \Rightarrow \frac{dx}{dy} = -\frac{e^{x/y}\left(1 - \frac{x}{y}\right)}{1 + e^{x/y}} \)
The right-hand side is a function of \( \frac{x}{y} \) only, i.e. a homogeneous function of degree zero, so the given equation is homogeneous.
Putting x = vy and \( \frac{dx}{dy} = v + y\frac{dv}{dy} \), we get
\( v + y\frac{dv}{dy} = -\frac{e^v(1 - v)}{1 + e^v} \)
⇒ \( y\frac{dv}{dy} = -\frac{e^v(1 - v)}{1 + e^v} - v \)
⇒ \( y\frac{dv}{dy} = \frac{-e^v + ve^v - v - ve^v}{1 + e^v} \)
⇒ \( y\frac{dv}{dy} = -\frac{v + e^v}{1 + e^v} \)
⇒ \( \frac{1 + e^v}{v + e^v}\,dv = -\frac{1}{y}\,dy \)
⇒ \( \int \frac{1 + e^v}{v + e^v}\,dv = -\int \frac{1}{y}\,dy \)
⇒ \( \log(v + e^v) = -\log y + \log C \)
⇒ \( v + e^v = \frac{C}{y} \Rightarrow \frac{x}{y} + e^{x/y} = \frac{C}{y} \Rightarrow x + ye^{x/y} = C \)