# Mathematics Questions and Answers – Methods of Solving First Order & First Degree Differential Equations

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This set of Mathematics 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

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

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

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

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

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)

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

Explanation: Given that, $$\frac{dy}{dx}=3e^x+2$$
Separating the variables, we get
dy=(3ex+2)dx
Integrating both sides, we get
$$\int dy=\int (3e^x+2)\,dx$$ –(1)
y=3ex+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

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=$$\frac{(2x-1)^2}{3}$$

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

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

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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