Question

Find the general solution for the equation (px-py)(py+x)=2p by reducing into Clairaut’s form by using the substitution X=x^2, Y=y^2  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) \(x^2 = y^2 + \frac{c}{2c+2}\)

I have been asked this question by my college director while I was bunking the class.

This intriguing question originated from Clairaut’s and Lagrange Equations topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (b) \(y^2 = cx^2 – \frac{2c}{c+1}\)

To explain I would say: \(X=x^2 \rightarrow \frac{dX}{dx} = 2x\)

\(Y=y^2 \rightarrow \frac{dY}{dy} = 2y\)

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 \(y^2 = cx^2 – \frac{2c}{c+1}\).