250+ TOP MCQs on Clairaut’s and Lagrange Equations and Answers

Ordinary Differential Equations Multiple Choice Questions 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
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=2sqrt{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)
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})
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) = 2sqrt{frac{X}{Y}} P)
(frac{(PX-Y)}{sqrt{Y}} (P+1)sqrt{X} = 2sqrt{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}} )
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=4pleft(frac{-1}{2p} + frac{c}{p^2}right) + log p)
(y(p) = frac{2c}{p} – 1 + frac{log⁡p}{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} )
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}.)

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