Ordinary Differential Equations Questions and Answers – Bernoulli Equations


This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Bernoulli Equations”.

1. Solution of the differential equation \(\frac{dy}{dx} + \frac{y}{x} = y^2\) x is ________
a) \(\frac{1}{y} = -x + c\)
b) \(\frac{1}{xy} = -x + c\)
c) \(\frac{1}{(xy)^2} = -y + c\)
d) \(\frac{1}{xy} = -x^2 + c\)
View Answer

Answer: b
Explanation: Given equation is of the form \(\frac{dy}{dx} + Py = Qy^n\) …divide by y2
where P and Q are functions of x hence this is Bernoulli’s equation in \(y, \frac{1}{y^2} \frac{dy}{dx} + \frac{1}{yx} = x\)
put \(\frac{1}{y} = t \rightarrow \frac{-1}{y^2} \frac{dy}{dx} = \frac{dt}{dx}\) substituting we get – \(\frac{dt}{dx} + \frac{t}{x} = x \,or\, \frac{dt}{dx} – \frac{t}{x} = -x\)
this equation is linear in t i.e it is of the form
\(\frac{dt}{dx} + Pt = Q, I.F = e^{\int P \,dx} =e^{\int \frac{-1}{x} \,dx} = e^{-log⁡x} = \frac{1}{x}\)
its solution is
\(te^{\int P \,dx} = \int Q e^{\int P \,dx} \,dx + c \rightarrow t \frac{1}{x} = \int -x * \frac{1}{x} \,dx + c \rightarrow \frac{t}{x} = -x + c \,but\, t = \frac{1}{y}\)
thus solution is given by \(\frac{1}{xy} = -x + c\).

2. Solution of the differential equation \(\frac{dy}{dx} -y \,tan⁡x = \frac{sin⁡x cos^2⁡x}{y^2}\) is ______
a) \(y \,cos^2 x = \frac{-cos^4 xsin^2 x}{2} + c\)
b) \(y^2 cos^2 x = \frac{sin^6 x}{2} + c \)
c) \(y^3 cos^3 x = \frac{-cos^6 x}{2} + c \)
d) \(y^4 cos^5 x = \frac{sinx cos⁡x}{2} + c \)
View Answer

Answer: c
Explanation: \(\frac{dy}{dx} -y \,tan⁡x = \frac{sin⁡x cos^2⁡x}{y^2}\) multiplying by
\(y^2 \rightarrow y^2 \frac{dy}{dx} – y^3 tan⁡x = sin⁡x \,cos^2⁡x \)
put \(t = y^3\rightarrow3y^2 \frac{dy}{dx} = \frac{dt}{dx} \,or\, y^2 \frac{dy}{dx} = \frac{1}{3} \frac{dt}{dx}\)
substituting \(\frac{dt}{dx}-3t \,tan⁡x = 3sin⁡x \,cos^2⁡x \)
this equation is linear in t i.e it is of the form \(\frac{dt}{dx} + Pt = Q, e^{\int P \,dx} = e^{\int -3tan⁡x \,dx}\)
\(e^{-3 log⁡\, sec⁡x} = cos^3 \,x\) its solution is \(te^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)
t cos3 x=∫3sin⁡x cos2x cos3 x dx + c = ∫3sin⁡x cos5⁡x dx+c, put v=cos x
dv=-sin x dx i.e \(\int 3sin⁡x \,cos^5\, ⁡x \,dx = \int -3v^5 \,dx = \frac{-v^6}{2} = \frac{-cos^6 x}{2}\) hence its solution becomes \(t cos^3 x = \frac{-cos^6 x}{2} + c \rightarrow y^3 cos^3 \,x = \frac{-cos^6 x}{2} + c.\)

3. Solution of the differential equation 6y2 dx – x(x3 + 2y)dy = 0 is ________
a) \(\frac{y}{x^3} = \frac{-log⁡y}{2} + c \)
b) \(\frac{y^2}{x^3} = \frac{-log⁡x}{4} + c \)
c) \(\frac{x}{y^3} = \frac{-log⁡x}{2} + c \)
d) \(\frac{x}{y^2} = \frac{-log⁡y}{4} + c \)
View Answer

Answer: a
Explanation: Equation is reduced to \(\frac{dy}{dx} = \frac{x(x^3+2y)}{6y^2} \,i.e\, \frac{dx}{dy}-\frac{x}{3y} = \frac{x^4}{6y^2}\) …divide by x4
we get \(\frac{1}{x^4} \frac{dy}{dx} – \frac{1}{3x^3 y} = \frac{1}{6y^2} \,put\, \frac{1}{x^3} = t \rightarrow \frac{-3}{x^4} \frac{dx}{dy} = \frac{dt}{dy} \,or\, \frac{1}{x^4} \frac{dy}{dx} = \frac{-1}{3} \frac{dt}{dy} \)
substituting \(\frac{-1}{3} \frac{dt}{dy} – \frac{t}{3y} = \frac{1}{6y^2} \,or\, \frac{dt}{dy} + \frac{t}{y} = \frac{-1}{2y^2}\) this equation is linear in t i.e it is of the form
\(\frac{dt}{dy} + Pt = Q \,where\, P=\frac{1}{y}, Q=\frac{-1}{2y^2} \,I.F\, = e^{\int P \,dy} = e^{\int \frac{1}{y} \,dy} = e^{log⁡y} = y\) its solution is \(te^{\int P \,dy} = \int Q \,e^{\int P \,dy} + c \rightarrow ty = \int y * \frac{-1}{2y^2} \,dy + c = \frac{-log⁡y}{2} + c\)
\(\frac{y}{x^3} = \frac{-log⁡y}{2} + c \) is the required solution.

Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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