Ordinary Differential Equations Questions and Answers – Clairaut’s and Lagrange Equations

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This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Clairaut’s and Lagrange Equations”.

1. Singular solution for the Clairaut’s equation \(y = y’x+\frac{a}{y’}\) is given by _______
a) \(\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1\)
b) y2=-4ax
c) y2=4ax
d) x2=-2ay
View Answer

Answer: c
Explanation: Let \( p = y’ = \frac{dy}{dx}\)
Given equation is of the form y = px + f(p), whose general solution is y = cx + f(c)
thus the general solution is \(y = cx + \frac{a}{c}\) …………..(1), to find the value of c
we differentiate (1) partially w.r.t ‘c’ i.e \( 0 = x-\frac{a}{c^2} \rightarrow c^2 = \frac{a}{x} \rightarrow c = \sqrt{\frac{a}{x}}\)
hence (1) becomes \(y = \sqrt{\frac{a}{x}} x + a \sqrt{\frac{x}{a}} \rightarrow y=2\sqrt{ax}\)
y2=4ax is the singular solution.
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2. Obtain the general solution for the equation xp2+px-py+1-y=0 where p=\(\frac{dy}{dx}\).
a) y=cx+\(\frac{1}{c+1}\)
b) x=cy-(c+1)
c) x=cy-\(\frac{1}{c+1}\)
d) y=cx+(c+1)
View Answer

Answer: a
Explanation: xp2+px-py+1-y=0
xp2+px+1=y(p+1)
\(y=\frac{xp(p+1)+1}{p+1} \,or\, y=px+\frac{1}{p+1}\)……(1) thus (1) is in the Clairaut’s equation form y=px+f(p),
thus general solution is y=cx+\(\frac{1}{c+1}\).

3. Find the general solution for the equation (px-py)(py+x)=2p by reducing into Clairaut’s form by using the substitution X=x2, Y=y2 where p=\(\frac{dy}{dx}\).
a) \(y^2 = x + \frac{c}{c+1}\)
b) \(y^2 = cx^2 – \frac{2c}{c+1}\)
c) \(x^2 = cy^2 – \frac{1}{2c+1}\)
d) \(x^2 = y^2 + \frac{c}{2c+2}\)
View Answer

Answer: b
Explanation: \(X=x^2 \rightarrow \frac{dX}{dx} = 2x\)
\(Y=y^2 \rightarrow \frac{dY}{dy} = 2y\)
now \(p = \frac{dy}{dx} = \frac{dy}{dY} \frac{dY}{dX} \frac{dX}{dx} \,and\, \,let\, P=\frac{dY}{dx}\)
\(p=\frac{1}{2y} * P * 2x \,or\, p=\frac{x}{y} \,P \,i.e\, p=\sqrt{\frac{X}{Y}} P\)
now consider (px-py)(py+x)=2p substituting the value of p we get
\(\left(\sqrt{\frac{X}{Y}} P \sqrt{X} – \sqrt{Y}\right)\left(\sqrt{\frac{X}{Y}} P \sqrt{Y} + \sqrt{X}\right) = 2\sqrt{\frac{X}{Y}} P\)
\(\frac{(PX-Y)}{\sqrt{Y}} (P+1)\sqrt{X} = 2\sqrt{\frac{X}{Y}} P \rightarrow (PX-Y)(P+1)=2P \,or\, Y(P)=PX-\frac{2P}{P+1}\) is in the Clairaut’s form
hence general solution is \(y^2 = cx^2 – \frac{2c}{c+1}\).

4. Find the general solution of the D.E 2y-4xy’-log y’=0.
a) \(y(p) = \frac{2c}{p} – 1 + \frac{log⁡p}{2} \)
b) \(y(p) = \frac{c}{2p} – 2 + log⁡p\)
c) \(x(p) = \frac{-1}{p} + \frac{c}{p^2} \)
d) \(x(p) = \frac{1}{2p} + \frac{c}{p^{1/2}} \)
View Answer

Answer: a
Explanation: Let y’=p and hence given equation is in Lagrange equation form
i.e 2y=4xp+log p …..(1) differentiating both sides of the equation
2dy=4xdp+4pdx+\(\frac{dp}{p}\) and dy=pdx
–> 2pdx=4xdp+4pdx+\(\frac{dp}{p}\) –> -2pdx=4pdx+\(\frac{dp}{p}\)
-2p\(\frac{dx}{dp} = 4x + \frac{1}{p} \rightarrow \frac{dx}{dp} + \frac{2}{p} x = \frac{-1}{2p^2}\) (p≠0)…….this is a linear D.E for the function x(p)
I.F is \(e^{\int \frac{2}{p} \,dp} = e^{log⁡p^2} = p^2\) and solution is x(p) \(p^2 = \int p^2 *\frac{-1}{2p^2} \,dp + c\)
\(x(p) = \frac{-1}{2p} + \frac{c}{p^2}\) substituting back in (1) we get \(2y=4p\left(\frac{-1}{2p} + \frac{c}{p^2}\right) + log p\)
\(y(p) = \frac{2c}{p} – 1 + \frac{log⁡p}{2} \).
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5. Find the general solution of the D.E y = 2xy’ – 3(y’)2.
a) \(y(p) = p^{1/2} + \frac{c}{2p}\)
b) \(y(p) = p^2 + \frac{2c}{p}\)
c) \(x(p) = -cp + \frac{c}{p^2}\)
d) \(x(p) = 2p + \frac{2c}{p^2} \)
View Answer

Answer: b
Explanation: Let y’=p –> y = 2xp – 3p2 ….(1) is in the Lagrange equation form
now differentiating we get dy=2xdp+2pdx-6pdp and dy=pdx
thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp-6pdp –> \(\frac{dx}{dp} + \frac{2}{p} x – 6=0\)…(2)
(2) is a linear D.E whose I.F=\(e^{\int \frac{2}{p} \,dp} = p^2\) hence its solution is
\(p^2 x(p) = \int 6p^2 \,dp + c \rightarrow x(p) = 2p + \frac{c}{p^2}\) ….substituting in (1) we get
\(y(p) = 2(2p+\frac{c}{p^2})p-3p^2 = p^2 + \frac{2c}{p}.\)

Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations, here is complete set of 1000+ Multiple Choice Questions and Answers.

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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