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Changing the Independent Variable in a DE | JEE

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Question
If the independent variable xx is changed to yy, then the differential equation xd2ydx2+(dydx)3−dydx=0x\dfrac{d^2y}{dx^2} + \left(\dfrac{dy}{dx}\right)^3 - \dfrac{dy}{dx} = 0 is transformed to xd2xdy2+(dxdy)2=kx\dfrac{d^2x}{dy^2} + \left(\dfrac{dx}{dy}\right)^2 = k, where kk is a number. Then kk equals:
A00
B11correct
C−1-1
D22
Solution
Use the change-of-variable formulas
dydx=1dxdy,d2ydx2=−d2xdy2(dxdy)3.\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}, \qquad \frac{d^2y}{dx^2} = -\frac{\frac{d^2x}{dy^2}}{\left(\frac{dx}{dy}\right)^3}.
 Substitute into xd2ydx2+(dydx)3−dydx=0x\dfrac{d^2y}{dx^2} + \left(\dfrac{dy}{dx}\right)^3 - \dfrac{dy}{dx} = 0 (write x′=dxdyx' = \frac{dx}{dy}, x′′=d2xdy2x'' = \frac{d^2x}{dy^2}):
−x x′′(x′)3+1(x′)3−1x′=0.-\frac{x\,x''}{(x')^3} + \frac{1}{(x')^3} - \frac{1}{x'} = 0.
 Multiply by −(x′)3-(x')^3:
x x′′−1+(x′)2=0⇒xd2xdy2+(dxdy)2=1.x\,x'' - 1 + (x')^2 = 0 \Rightarrow x\frac{d^2x}{dy^2} + \left(\frac{dx}{dy}\right)^2 = 1.
 ∴k=1\therefore k = 1. Correct answer: (2)
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