Particular Solutions of Differential Equation - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “General and Particular Solutions of Differential Equation”.

1. Which of the following functions is the solution of the differential equation \(\frac{dy}{dx}\)+2y=0?
a) y=-2e^-x
b) y=2e^x
c) y=e^-2x
d) y=e^2x
View Answer

Answer: c
Explanation: Consider the function y=e^-2x
Differentiating both sides w.r.t x, we get
\(\frac{dy}{dx}=-2e^{-2x}\)
\(\frac{dy}{dx}\)=-2y
⇒\(\frac{dy}{dx}\)+2y=0.

2. The function y=8 sin⁡2x is a solution of the differential equation \(\frac{d^2 y}{dx^2}\)+4y=0.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true.
Consider the function y=8 sin⁡2x
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=16 cos⁡2x –(1)
Differentiating (1) w.r.t x, we get
\(\frac{d^2 y}{dx^2}\)=-32 sin⁡2x
\(\frac{d^2 y}{dx^2}\)=-4(8 sin⁡2x )=-4y
⇒\(\frac{d^2 y}{dx^2}\)+4y=0.

3. Which of the following functions is a solution for the differential equation xy’-y=0?
a) y=4x
b) y=x²
c) y=-4x
d) y=2x
View Answer

Answer: d
Explanation: Consider the function y=2x
Differentiating w.r.t x, we get
y’=\(\frac{dy}{dx}\)=2
Substituting in the equation xy’-y, we get
xy’-y=x(2)-2x=2x-2x=0
Therefore, the function y=2x is a solution for the differential equation xy’-y=0.

4. Which of the following differential equations has the solution y=3x²?
a) \(\frac{d^2 y}{dx^2}\)-6x=0
b) \(\frac{dy}{dx}\)-3x=0
c) x \(\frac{d^2 y}{dx^2}\)–\(\frac{dy}{dx}\)=0
d) \(\frac{d^2 y}{dx^2}-\frac{3dy}{dx}\)=0
View Answer

Answer: c
Explanation: Consider the function y=3x²
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=6x –(1)
Differentiating (1) w.r.t x, we get
\(\frac{d^2 y}{dx^2}\)=6
∴\(\frac{xd^2 y}{dx^2}-\frac{6dy}{dx}\)=6x-6x=0
Hence, the function y=3x² is a solution for the differential equation x \(\frac{d^2 y}{dx^2}\)-6 \(\frac{dy}{dx}\)=0.

5. Which of the following functions is a solution for the differential equation y”+6y=0?
a) y=5 cos⁡3x
b) y=5 tan⁡3x
c) y=cos⁡3x
d) y=6 cos⁡3x
View Answer

Answer: a
Explanation: Consider the function y=5 cos⁡3x
Differentiating w.r.t x, we get
y’=\(\frac{dy}{dx}\)=-15 sin⁡3x
Differentiating again w.r.t x, we get
y”=\(\frac{d^2 y}{dx^2}\)=-30 cos⁡3x
⇒y”+6y=0.
Hence, the function y=5 cos⁡3x is a solution for the differential equation y”+6y=0.

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6. Which of the following functions is a solution for the differential equation \(\frac{dy}{dx}\)-14x=0?
a) y=7x²
b) y=7x³
c) y=x⁷
d) y=14x
View Answer

Answer: a
Explanation: Consider the function y=7x²
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=14x
∴\(\frac{dy}{dx}\)-14x=0
Hence, the function y=7x² is a solution for the differential equation \(\frac{dy}{dx}\)-14x=0

7. Which of the following differential equations given below has the solution y=log⁡x?
a) \(\frac{d^2 y}{dx^2}\)-x=0
b) \(\frac{d^2 y}{dx^2}+(\frac{dy}{dx})^2\)=0
c) \(\frac{d^2 y}{dx^2}\)–\(\frac{dy}{dx}\)=0
d) x \(\frac{d^2 y}{dx^2}\)-log⁡x=0
View Answer

Answer: b
Explanation: Consider the function y=log⁡x
Differentiating w.r.t x, we get
\(\frac{dy}{dx}=\frac{1}{x} \)–(1)
Differentiating (1) w.r.t x, we get
\(\frac{d^2 y}{dx^2}=-\frac{1}{x^2} \)
∴\(\frac{d^2 y}{dx^2}+(\frac{dy}{dx})^2=-\frac{1}{x^2}+(\frac{1}{x})^2\)
=-\(\frac{1}{x^2}+\frac{1}{x^2}\)=0.

8. How many arbitrary constants will be there in the general solution of a second order differential equation?
a) 3
b) 4
c) 2
d) 1
View Answer

Answer: c
Explanation: The number of arbitrary constants in a general solution of a n^th order differential equation is n.
Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.

9. The number of arbitrary constants in a particular solution of a fourth order differential equation is __________________
a) 1
b) 0
c) 4
d) 3
View Answer

Answer: b
Explanation: The number of arbitrary constants for a particular solution of n^th order differential equation is always zero.

10. The function y=3 cos⁡x is a solution of the function \(\frac{d^2 y}{dx^2}-3\frac{dy}{dx}\)=0.
a) True
b) False
View Answer

Answer: b
Explanation: The given statement is false.
Given differential equation: \(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=0 –(1)
Consider the function y=3 cos⁡x
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=-3 sin⁡x
Differentiating again w.r.t x, we get
\(\frac{d^2 y}{dx^2}\)=-3 cos⁡x
Substituting the values of \(\frac{dy}{dx}\) and \(\frac{d^2 y}{dx^2}\) in equation (1), we get
\(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=-3 cos⁡x-3(-3 sin⁡x)
=9 sin⁡x-3 cos⁡x≠0.
Hence, y=3 cos⁡x, is not a solution of the equation \(\frac{d^2 y}{dx^2}\)-3 \(\frac{dy}{dx}\)=0.

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