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
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.
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
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
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) \(y(p) = \frac{2c}{p} – 1 + \frac{logp}{2} \)
b) \(y(p) = \frac{c}{2p} – 2 + logp\)
c) \(x(p) = \frac{-1}{p} + \frac{c}{p^2} \)
d) \(x(p) = \frac{1}{2p} + \frac{c}{p^{1/2}} \)
View Answer
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 \(e^{\int \frac{2}{p} \,dp} = e^{logp^2} = p^2\) 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 \(2y=4p\left(\frac{-1}{2p} + \frac{c}{p^2}\right) + log p\)
\(y(p) = \frac{2c}{p} – 1 + \frac{logp}{2} \).
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
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.
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