Engineering Questions with Answers - Multiple Choice Questions

Ordinary Differential Equations MCQ – Clairaut’s and Lagrange Equations

1 - Question

Singular solution for the Clairaut’s equation y=yx+ay is given by _______
a) x2a2+y2a2=1
b) y2=-4ax
c) y2=4ax
d) x2=-2ay

View Answer

Answer: c
Explanation: Let p=y=dydx
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+ac …………..(1), to find the value of c
we differentiate (1) partially w.r.t ‘c’ i.e 0=xac2c2=axc=ax−−√
hence (1) becomes y=ax−−√x+axa−−√y=2ax−−√
y2=4ax is the singular solution.

2 - Question

 Obtain the general solution for the equation xp2+px-py+1-y=0 where p=dydx.
a) y=cx+1c+1
b) x=cy-(c+1)
c) x=cy-1c+1
d) y=cx+(c+1)

View Answer

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

3 - Question

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=dydx.
a) y2=x+cc+1
b) y2=cx22cc+1
c) x2=cy212c+1
d) x2=y2+c2c+2

View Answer

Answer: b
Explanation: X=x2dXdx=2x
now p=dydx=dydYdYdXdXdxandletP=dYdx
now consider (px-py)(py+x)=2p substituting the value of p we get
(PXY)Y(P+1)X−−√=2XY−−√P(PXY)(P+1)=2PorY(P)=PX2PP+1 is in the Clairaut’s form
hence general solution is y2=cx22cc+1.

4 - Question

Find the general solution of the D.E 2y-4xy’-log y’=0.
a) y(p)=2cp1+logp2
b) y(p)=c2p2+logp
c) x(p)=1p+cp2
d) x(p)=12p+cp1/2

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+dpp and dy=pdx
–> 2pdx=4xdp+4pdx+dpp –> -2pdx=4pdx+dpp
-2pdxdp=4x+1pdxdp+2px=12p2 (p≠0)…….this is a linear D.E for the function x(p)
I.F is e2pdp=elogp2=p2 and solution is x(p) p2=p212p2dp+c
x(p)=12p+cp2 substituting back in (1) we get 2y=4p(12p+cp2)+logp

5 - Question

Find the general solution of the D.E y = 2xy’ – 3(y’)2.
a) y(p)=p1/2+c2p
b) y(p)=p2+2cp
c) x(p)=cp+cp2
d) x(p)=2p+2cp2

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 –> dxdp+2px6=0…(2)
(2) is a linear D.E whose I.F=e2pdp=p2 hence its solution is
p2x(p)=6p2dp+cx(p)=2p+cp2 ….substituting in (1) we get

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