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
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)
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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)
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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=3ex+2x+C
b) y=3ex-2x+C
c) y=2ex+3x+C
d) y=2ex-3x+C
View Answer

Answer: a
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
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.
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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
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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.

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

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter