Methods of Solving First Order and Degree Differential Equations - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “Methods of Solving First Order & First Degree Differential Equations”.

1. Find the general solution of the differential equation \(\frac{dy}{dx}=5x^2+2\).
a) 10x³+12x-3y²+C=0
b) 12x-3y²+C=0
c) 10x³+12x-y²+C=0
d) 10x²-3y²+C=0
View Answer

Answer: a
Explanation: Given that, \(\frac{dy}{dx}=5x^2+2\)
Separating the variables, we get
dy=(5x²+2)dx –(1)
Integrating both sides of (1), we get
\(\int y \,dy=\int 5x^2+2 \,dx\)
\(\frac{y^2}{2}=\frac{5x^3}{3}+2x+C_1\)
3y²=\(10x^3+12x+6C_1\)
10x³+12x-3y²+C=0 (where 6C₁=C)

2. Find the general solution of the differential equation \(\frac{dy}{dx}=\frac{y-3}{x-3}\) (x, y≠3).
a) x-3=0
b) y-3=0
c) y+3=0
d) x-3y=0
View Answer

Answer: b
Explanation: Given that, \(\frac{dy}{dx}=\frac{y-3}{x-3}\)
Separating the variables, we get
\(\frac{dy}{y-3}=\frac{dx}{x-3}\)
log⁡(y-3)=log⁡(x-3)+log⁡C₁
log⁡(y-3)-log⁡(x-3)=log⁡C₁
\(log⁡(\frac{y-3}{x-3})\)=log⁡C₁
\(\frac{1}{C_1} \frac{y-3}{x-3}=0\)
y-3=0 is the general solution for the given differential equation.

3. Find the general solution of the differential solution \(\frac{dy}{dx}=2-x+x^3\).
a) x⁴-2x²-4y+C=0
b) x⁴-2x²+C=0
c) 2x²+4x-4y+C=0
d) x⁴-2x²+4x-4y+C=0
View Answer

Answer: d
Explanation: Given that, \(\frac{dy}{dx}\)=2-x+x⁴
Separating the variables, we get
dy=(2-x+x³)dx
Integrating on both sides, we get
\(\int dy=\int 2-x+x^3 \,dx\)
\(y=2x-\frac{x^2}{2}+\frac{x^4}{4}+C_1\)
4y=4x-2x²+x⁴+4C₁
∴x⁴-2x²+4x-4y+4C₁=0
x⁴-2x²+4x-4y+C=0 (where 4C₁=C)

4. Find the general solution of the differential equation \(\frac{dy}{dx}=\frac{3 \,sec\,⁡y}{2 \,cosec⁡\,x}\).
a) 3 cos⁡x-2 cos⁡y=C
b) 3 sin⁡x+2 sin⁡y=C
c) 3 cos⁡x+2 tan⁡x=C
d) 3 cos⁡x+2 sin⁡y=C
View Answer

Answer: d
Explanation: Given that, \(\frac{dy}{dx}=\frac{3 \,sec⁡\,y}{2cosec \,x}\)
\(\frac{2 \,dy}{sec⁡ \,y}=\frac{3dx}{cosec\,⁡x}\)
Separating the variables, we get
2 cos⁡y dy=3 sin⁡x dx
Integrating both sides, we get
∫ 2 cos⁡y dy = ∫ 3 sin⁡x dx
2 sin⁡y=3(-cos⁡x)+C
3 cos⁡x+2 sin⁡y=C

5. Find the general solution of the differential equation \(\frac{dy}{dx}=\frac{2+x^3}{4-y^3}\).
a) x³-y³-4y+C=0
b) x⁴+8x+y⁴-16y+C=0
c) 2x+y⁴-4y+C=0
d) x³+2x+C=0
View Answer

Answer: b
Explanation: Given that, \(\frac{dy}{dx}=\frac{2+x^3}{4-y^3}\)
Separating the variables, we get
(4-y³)dy=(2+x³)dx
Integrating both sides, we get
\(\int 4-y^3 \,dy=\int 2+x^3 \,dx\)
\(4y-\frac{y^4}{4}=2x+\frac{x^4}{4}+C_1\)
x⁴+8x+y⁴-16y+C=0 (where 4C₁=C)

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6. Find the general solution of the differential equation \(\frac{dy}{dx}=3e^x+2\)
a) y=3e^x+2x+C
b) y=3e^x-2x+C
c) y=2e^x+3x+C
d) y=2e^x-3x+C
View Answer

Answer: a
Explanation: Given that, \(\frac{dy}{dx}=3e^x+2\)
Separating the variables, we get
dy=(3e^x+2)dx
Integrating both sides, we get
\(\int dy=\int (3e^x+2)\,dx\) –(1)
y=3e^x+2x+C which is the general solution of the given differential equation.

7. Find the particular solution of the differential equation \(\frac{dy}{dx}\)+2x=5 given that y=5, when x=1.
a) y=5x+x²+1
b) y=x-x²+4
c) y=5x-x²+1
d) y=5x-x²
View Answer

Answer: c
Explanation: Given that, \(\frac{dy}{dx}+2x=5\)
\(\frac{dy}{dx}=5-2x\)
Separating the variables, we get
dy=(5-2x)dx
Integrating both sides, we get
\(\int dy=\int 5-2x \,dx\)
y=5x-x²+C –(1)
Given that, y=5, when x=1
⇒5=5(1)-(1)²+C
∴C=1
Substituting value of C to equation (1), we get
y=5x-x²+1 which is the particular solution of the given differential equation.

8. Find the particular solution of the differential equation \(\frac{dy}{dx}+8x=16x^2+4\) given that y=\(\frac{1}{3}\) when x=1.
a) y=\(\frac{(2x+1)^2}{3}\)
b) y=\(\frac{(4x+1)^2}{12}\)
c) y=\(\frac{(4x-2)^2}{3}\)
d) y=\(\frac{(2x-1)^2}{3}\)
View Answer

Answer: d
Explanation: Given that, \(\frac{dy}{dx}+8x=16x^2+4\)
\(\frac{dy}{dx}=16x^2-8x+4\)
\(\frac{dy}{dx}=(4x-2)^2\)
Separating the variables, we get
dy=(4x-2)² dx
Integrating both sides, we get
\(\int dy=\int (4x-2)^2 \,dx\)
y=\(\frac{(4x-2)^2}{12}+C\)
y=\(\frac{(2x-1)^2}{3}+C\) –(1)
Given that, y=1/3 when x=1
Therefore, equation (1) becomes,
\(\frac{1}{3}=\frac{(2(1)-1)^2}{3}+C\)
\(C=\frac{1}{3}-\frac{1}{3}\)=0
Hence, the particular solution for the given differential solution is y=\(\frac{(2x-1)^2}{3}\).

9. Find the particular solution for the differential equation \(\frac{dy}{dx}=\frac{3x^2}{7y}\) given that, y=1 when x=1.
a) 7x²=2y³+5
b) 7x³=2y²+5
c) 7y²=2x³+5
d) 2y²=5x³+6
View Answer

Answer: c
Explanation: Given that, \(\frac{dy}{dx}=\frac{3x^2}{7y}\)
Separating the variables, we get
7y dy=3x² dx
Integrating both sides, we get
\(7\int y \,dy=3\int x^2 \,dx\)
\(\frac{7y^2}{2}=3(\frac{x^3}{3})+C\)
\(\frac{7y^2}{2}=x^3\)+C –(1)
Given that y=1, when x=1
Substituting the values in equation (1), we get
\(\frac{7(1)^2}{2}=(1)^3+C\)
\(C=\frac{7}{2}-1=\frac{5}{2}\)
Hence, the particular solution of the given differential equation is:
\(\frac{7y^2}{2}=x^3+\frac{5}{2}\)
⇒7y²=2x³+5

10. Find the particular solution of the differential equation \(\frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}\).
a) (log⁡y)²+(log⁡x)²=0
b) (log⁡y)²-(log⁡x)²=0
c) log⁡y-log⁡x=0
d) 2 log⁡x+log⁡y=0
View Answer

Answer: b
Explanation: \(\frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}\)
Separating the variables, we get
\(\frac{5 \,log⁡y}{y} dy=\frac{9 \,log⁡x}{x} dx\)
Integrating both sides, we get
\(5\int \frac{log⁡y}{y} \,dy=9\int \frac{log⁡x}{x} \,dx\) –(1)
First, for integrating \(\frac{log⁡y}{y}\)
Let log⁡y=t
Differentiating w.r.t y, we get
\(\frac{1}{y}\) dy=dt
∴\(\int \frac{log⁡y}{y} \,dy=\int t \,dt\)
=\(\frac{t^2}{2}=\frac{log⁡y^2}{2}\)
Similarly integrating \(\frac{log⁡x}{x}\)
Let log⁡x=t
Differentiating w.r.t x, we get
\(\frac{1}{x}\) dx=dt
∴\(\int \frac{log⁡x}{x} dy=\int t \,dt\)
=\(\frac{t^2}{2}=\frac{(log⁡x)^2}{2}\)
Hence, equation (1), becomes
\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\) –(2)
Given y=2, we get x=2
Substituting the values in equation (2), we get
\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\)
C=0
Therefore, the equation becomes \((log⁡y)^2=(log⁡x^2)\)
∴(log⁡y)²-(log⁡x)²=0.

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