Answer to Question #139609 in Differential Equations for LAVANYA

Question

p^2(x^2-a^2)-2xyp+y^2+a^4=0 find G.S and S.S

Answer 1

Given differential equation is, $p^2(x^2-a^2)-2xyp+y^2+a^4=0$

I can write it as,

$p^2x^2-2xyp+y^2=a^2p^2 - a^4$

$\implies (y-xp)^2 = a^2(p^2-a^2)$

Then, $(y-xp) = \sqrt{a^2(p^2-a^2)}$

$y=xp \pm \sqrt{a^2(p^2-a^2)}$

This equation is same as charpit's equation, $y = xf(p)+\phi(p)$

So general solution of the equation will be,

$y=cx \pm \sqrt{a^2(c^2-a^2)}$

where c is any constant.

For particular solution, we need some boundary conditions.

Let x = 0 at y=0

then,

$0=0c \pm \sqrt{a^2(c^2-a^2)} \implies c = \pm a$

so, particular solution will be,

$y=\pm ax$