Answer to Question #185541 in Differential Equations for John

Question

p(p+y)=x(x+y)

Answer 1

We have given the differential equation,

$p(p+y) = x(x+y)$

$p^2 + py - x(x+y) = 0$

$p =\dfrac{-y \pm \sqrt{y^2+4x(x+y)}}{2}$

$p = \dfrac{-y \pm (2x+y)}{2}$

Taking the positive sign,

$p = \dfrac{-y+2x+y}{2}$

$p = x$

$\dfrac{dy}{dx} = x$

$dy = xdx$

$y = \dfrac{x^2}{2}+C$ ---------- (First solution)

Now, Taking the negative sign.

$p = \dfrac{-y-2x-y}{2}$

$p = -(x+y)$

$\dfrac{dy}{dx} = -x-y$

$\dfrac{dy}{dx}+y = -x$

This is a linear first order differential equation.

$I.F = e^x$

Its solution can be given as,

$ye^x = \int -xe^x +C$

$ye^x = -(x-1)e^x + C$

$y = -(x-1) + Ce^{-x}$ ------- (Second solution)