Class 12 Differential Equations Mock Test (CBSE)

Topics in Differential Equations

• Order and degree of a differential equation
• General and particular solutions
• Variable separable method
• Homogeneous differential equations
• Linear differential equations

Differential Equations practice questions with answers

1. Q1Multiple choice·1 mark
   Which substitution reduces the homogeneous differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ to variable separable form?

   1. (a)$v = x + y$
   2. (b)$v = x - y$
   3. (c)$x = vy^2$
   4. (d)$y = vx$
   Show answer and step marks

   1. Correct option: (d) $y = vx$, which gives $\frac{dy}{dx} = v + x\frac{dv}{dx}$ and the equation becomes $x\frac{dv}{dx} = \frac{1}{v}$.+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   Which of the following is a homogeneous differential equation?

   1. (a)$\frac{dy}{dx} = \frac{x^2 + 1}{y}$
   2. (b)$\frac{dy}{dx} = \frac{x + y + 1}{x - y}$
   3. (c)$\frac{dy}{dx} = \frac{x^2 + 3xy}{y^2}$
   4. (d)$\frac{dy}{dx} = \frac{x^2 + y}{xy}$
   Show answer and step marks

   1. Correct option: (c) $\frac{dy}{dx} = \frac{x^2 + 3xy}{y^2}$, since replacing $x$ by $\lambda x$ and $y$ by $\lambda y$ gives $\frac{\lambda^2(x^2+3xy)}{\lambda^2 y^2}$, which is unchanged.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   The degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^3 + 4\left(\frac{dy}{dx}\right)^2 + y = 0$ is:

   1. (a)3
   2. (b)Not defined
   3. (c)1
   4. (d)2
   Show answer and step marks

   1. Correct option: (a) $3$+1

   Answer: (a)
4. Q4Assertion and reason·1 mark
   Assertion (A): The differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$ is homogeneous. Reason (R): Every differential equation of first order and first degree is homogeneous.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (c) Assertion (A) is true but Reason (R) is false, since $\frac{dy}{dx} = \frac{1 + y/x}{1 - y/x}$ is homogeneous, while many first order first degree equations such as $\frac{dy}{dx} = x + y$ are not.+1

   Answer: (c)
5. Q5Short answer·2 marks
   Verify that the function $y = a \cos x + b \sin x$, where $a, b \in \mathbb{R}$, is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.
   Show answer and step marks

   1. Find the first and second derivatives of $y$ with respect to $x$: $\frac{dy}{dx} = -a \sin x + b \cos x$ and $\frac{d^2y}{dx^2} = -a \cos x - b \sin x$.+1
   2. Substitute the expression for $\frac{d^2y}{dx^2}$ and $y$ into the L.H.S. of the differential equation: $(-a \cos x - b \sin x) + (a \cos x + b \sin x) = 0 = \text{R.H.S}$. Hence verified.+1

   Answer: $\frac{d^2y}{dx^2} = -a\cos x - b\sin x = -y$, so $\frac{d^2y}{dx^2} + y = 0$; hence $y = a\cos x + b\sin x$ is a solution.
6. Q6Short answer·3 marks
   A water tank at a guesthouse in a hill station holds $y$ litres of water at time $t$ minutes. Water is pumped in at a constant rate of 4 litres per minute, while water leaks out at a rate of $\frac{y}{20}$ litres per minute. At $t = 0$, the tank holds 10 litres. Find $y$ in terms of $t$.
   Show answer and step marks

   1. Rate balance gives $\frac{dy}{dt} = 4 - \frac{y}{20}$, i.e. $\frac{dy}{dt} + \frac{1}{20}y = 4$, which is linear in $y$.+0.5
   2. Integrating factor $= e^{\int \frac{1}{20}dt} = e^{t/20}$.+0.5
   3. Multiplying by the integrating factor, $\frac{d}{dt}\left(y e^{t/20}\right) = 4e^{t/20}$, so $y e^{t/20} = 80e^{t/20} + C$.+1
   4. Using $y = 10$ at $t = 0$ gives $C = -70$, so $y = 80 - 70e^{-t/20}$.+1

   Answer: $y = 80 - 70e^{-t/20}$

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