Answer to Question #211199 in Differential Equations for AMULYA HS

Question

find general and singular solution y=2xp+4yp^2;X=x,Y=y^2 ,by using clairat's equation

Answer 1

$y=2xp+4yp^2$

Multiplying by y, we obtain

$y^2=x\cdot 2yp+4y^2p^2$

By using variables $X=x$ , $Y=y^2$ , $P=dY/dX=2ydy/dx=2yp$ , we can re-write this equation as

$Y=XP+P^2$

This is Clairaut's equation. Differentiate it by X:

$dY/dX=P=P+XP'+2PP'$

$P'(X+2P)=0$

The general solution: $P'=0$ , $P=C$,

$Y=XP+P^2=CX+C^2$ ,

$y^2=Cx+C^2$

The singular solution: $X+2P=0$ , $P=-X/2$ , $Y=XP+P^2=X(-X/2)+(-X/2)^2=-X^2/4$ ,

$y^2=-x^2/4$

$y=\pm ix/2$ (acceptable, if x, y take values in complex numbers).

Answer. The general solution: $y^2=Cx+C^2$. The singular solution:

$y=\pm ix/2$ (acceptable, if x, y take values in complex numbers).