Question

x + 2(xp − y) + p^2 = 0

Answer 1

Solution;

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

$x+2xp-2y+p^2=0$

$2y=x+2xp+p^2$ .....(1)

$2y=x(1+2p)+p^2$ ....(1a)

Differentiate (1) with respect forms;

$2\frac{dy}{dx}=2p=1+2p+2x\frac{dp}{dx}+2p\frac{dp}{dx}$

Resolve as;

$0=1+2\frac{dp}{dx}(x+p)$

$\frac{dp}{dx}=\frac{-1}{2(x+p)}$

$\frac{dx}{dp}=-2(x+p)$

$\frac{dx}{dp}+2x=-2p$

The I.F;

$I.F=e^{\int{2}dp}=e^{2p}$

Hence;

$x×I.F=\int-2pe^{2p}$

$xe^{2p}=\frac{-(2p-1)e^{2p}}{2}+c$

Therefore;

$x=\frac{1-2p}{2}+\frac{c}{e^{2p}}$ ....(3)

Substitute (2) into (1a);

$2y=[\frac{1-2p}{2}+\frac{c}{e^{2p}}](1+2p)+p^2$

$y=[\frac{1-2p}{2}+\frac{c}{e^{2p}}]\frac{(1+2p)}{2}+\frac{p^2}{2}$