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=y′x+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=x−ac2→c2=ax→c=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
xp2+px+1=y(p+1)
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=cx2–2cc+1
c) x2=cy2–12c+1
d) x2=y2+c2c+2
View Answer
Answer: b
Explanation: X=x2→dXdx=2x
Y=y2→dYdy=2y
now p=dydx=dydYdYdXdXdxandletP=dYdx
p=12y∗P∗2xorp=xyPi.ep=XY−−√P
now consider (px-py)(py+x)=2p substituting the value of p we get
(XY−−√PX−−√–Y−−√)(XY−−√PY−−√+X−−√)=2XY−−√P
(PX−Y)Y√(P+1)X−−√=2XY−−√P→(PX−Y)(P+1)=2PorY(P)=PX−2PP+1 is in the Clairaut’s form
hence general solution is y2=cx2–2cc+1.
4 - Question
Find the general solution of the D.E 2y-4xy’-log y’=0.
a) y(p)=2cp–1+logp2
b) y(p)=c2p–2+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+1p→dxdp+2px=−12p2 (p≠0)…….this is a linear D.E for the function x(p)
I.F is e∫2pdp=elogp2=p2 and solution is x(p) p2=∫p2∗−12p2dp+c
x(p)=−12p+cp2 substituting back in (1) we get 2y=4p(−12p+cp2)+logp
y(p)=2cp–1+logp2.
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+2px–6=0…(2)
(2) is a linear D.E whose I.F=e∫2pdp=p2 hence its solution is
p2x(p)=∫6p2dp+c→x(p)=2p+cp2 ….substituting in (1) we get
y(p)=2(2p+cp2)p−3p2=p2+2cp.