250+ TOP MCQs on Methods of Solving First Order & First Degree Differential Equations | Class 12 Maths

Mathematics Multiple Choice Questions 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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)

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
Answer: a
Clarification: 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
Answer: c
Clarification: 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})
Answer: d
Clarification: 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
Answer: c
Clarification: Given that, (frac{dy}{dx}=frac{3x^2}{7y})
Separating the variables, we get
7y dy=3x² dx
Integrating both sides, we get
(7int y ,dy=3int 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
Answer: b
Clarification: (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
(5int frac{log⁡y}{y} ,dy=9int 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.