Question

Obtain the general solution for the equation xp^2+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)

This question was posed to me in an international level competition.

This is a very interesting question from Clairaut’s and Lagrange Equations topic in chapter Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) y=cx+\(\frac{1}{c+1}\)

To explain: xp^2+px-py+1-y=0

xp^2+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+\(\frac{1}{c+1}\).