Ordinary Differential Equations Questions and Answers for Freshers

This set of Ordinary Differential Equations Questions and Answers for Freshers focuses on “First Order Linear Differential Equations”.

1. Solution of the differential equation \(\frac{dy}{dx}\) + y cot ⁡x = cos⁡x is ______
a) \(y cos \,x = \frac{sin^2 x}{2} + c\)
b) \(y sin \,x = \frac{sin^2 x}{2} + c\)
c) \(y sin \,x = \frac{cos^2 x}{4} + c\)
d) \(y cos \,x = -\frac{sin^2 x}{4} + c\)
View Answer

Answer: b
Explanation: \(\frac{dy}{dx}\) + y cot⁡x = cos⁡x is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x only
given DE is linear DE in y here P=cot x, Q=cos x, \(e^{\int P \,dx} = e^{\int cot⁡x \,dx} = e^{log⁡sin⁡x} = sin x\)
Linear DE solution is given by \(y e^{\int P \,dx} = \int Q \,e^{\int P \,dx} \,dx + c\)
y sin x=∫(cos x*sin x) dx + c….. substitute sin x=t to solve integral
\(y sin x = \frac{t^2}{2} + c = \frac{sin^2 x}{2} + c\) is the solution.

2. For the differential equation \(\frac{dy}{dx}\) – 3y cot⁡x = sin⁡2x; y=2 when x=\(\frac{\pi}{2}\), its particular solution is ______
a) y = 2cos² x + 4cos³ x
b) y = -2sin³ x + 4sin² x
c) y = -2sin² x + 4sin³ x
d) y = 4cos² x + 2sin³ x
View Answer

Answer: c
Explanation: \(\frac{dy}{dx}\) – 3y cot⁡x = sin⁡2x is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x
only given DE is linear DE in y here P=-3cot x, Q=sin 2x, \(e^{\int P \,dx} = e^{\int -3cot⁡x \,dx}\)
\( =e^{-3 log⁡sin⁡x} = e^{log⁡(\frac{1}{sin^3x})} = \frac{1}{sin^3x}\)
Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)
\(y \frac{1}{sin^3x} = \int sin \,2x \frac{1}{sin^3x} \,dx + c = \int 2 \,sin⁡x \,cos⁡x \frac{1}{sin^3x} dx = \int 2 \,cos⁡x \frac{1}{sin^2x} dx\)
substitute sin x=t to solve integral \(y \frac{1}{sin^3x} = \int 2 \frac{1}{t^2} \,dt + c = \frac{-2}{t} + c = \frac{-2}{sin⁡x} + c\)
it is given that when x=π/2, y=2→2=-2+c→c=4 its particular solution is thus given
by \(y \frac{1}{sin^3x} = \frac{-2}{sin⁡x} + 4 \rightarrow y = -2sin^2 \,x + 4sin^3 \,x.\)

3. Solution of the differential equation \((x+3y^2)\frac{dy}{dx} = y(y>0)\) is ______________
a) x=3y²+cy
b) y=2x²+c
c) x=2y²+\(\frac{c}{y}\)
d) y=3x³+c
View Answer

Answer: a
Explanation: \((x+3y^2)\frac{dy}{dx} = y\) can be rearranged to \(\frac{dx}{dy} = \frac{x}{y} + 3y \rightarrow \frac{dx}{dy} – \frac{x}{y} = 3y \)
above equation is of the form \(\frac{dx}{dy}\) + Px = Q where P & Q is a function of y only
given DE is linear DE in x here\( P = \frac{-1}{y}, Q=3y, e^{\int P \,dy} = e^{\int \frac{-1}{y} \,dy} = e^{log⁡y^{-1}} = \frac{1}{y}\)
therefore its solution is given by \(xe^{\int P \,dy} = \int Q \,e^{\int P \,dy} dy + c\)
\(x \frac{1}{y} = \int 3y * \frac{1}{y} \,dy + c = 3y + c\)
i.e x=3y²+cy.

4. Solution of the differential equation \(\Big[\frac{e^{-2\sqrt{x}}}{\sqrt{x}} – \frac{y}{\sqrt{x}}\Big] \frac{dy}{dx} = 1, \,x≠0\) is _______
a) \(ye^{2\sqrt{x}}=2\sqrt{x}+c\)
b) \(ye^{-2\sqrt{x}}=2x^{-\frac{3}{2}}+c\)
c) \(ye^{-2\sqrt{x}}=3\sqrt{x}+c\)
d) \(ye^{2\sqrt{x}}=3x^{\frac{3}{2}}+c\)
View Answer

Answer: a
Explanation: \( \frac{1}{\sqrt{x}} e^{-2\sqrt{x}} – \frac{1}{\sqrt{x}} y = \frac{dy}{dx} \rightarrow \frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{1}{\sqrt{x}} e^{-2\sqrt{x}}\) is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x only.
given DE is linear DE in y here \(P=\frac{1}{\sqrt{x}}, Q=\frac{1}{\sqrt{x}} e^{-2\sqrt{x}}\)
\(e^{\int P \,dx} = e^{∫ \frac{1}{\sqrt{x}} \,dx} = e^{2\sqrt{x}},\)
Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)
\(ye^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} e^{-2\sqrt{x}} e^{2\sqrt{x}} \,dx + c = 2\sqrt{x} + c\) is the solution.

5. Particular solution of the differential equation \(\frac{dy}{dx} = \frac{x+y}{x}:y(1)=1\) is _____
a) x =y log|x| + y
b) y = y log|x| + 2x
c) x = x log|y| + y
d) y = x log|x| + x
View Answer

Answer: d
Explanation: \(\frac{dy}{dx} – \frac{y}{x} = 1\) is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x only
given DE is linear DE in y here \(P=\frac{-1}{x}, Q=1, e^{\int P \,dx} = e^{\int \frac{-1}{x} \,dx} = e^{logx^{-1}} = \frac{1}{x}\)
Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c \rightarrow y \frac{1}{x} = \int 1 \frac{1}{x} \,dx + c\)
\(\frac{y}{x}\) = log⁡|x|+c, given y(1)=1–>y=1 when x=1 i.e c=1 therefore its particular solution is given by y = x log|x| + x.

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