Ordinary Differential Equations Questions and Answers – Bernoulli Equations

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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 \(\frac{1}{xy} = -x + c\).
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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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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