Answer to Question #208335 in Differential Equations for cglgl

Question

Solve xp2-yp-y=0

Answer 1

Solve

$$xp^2-yp-y=0, p=\dfrac{dy}{dx}=y'$$

Solve for $y$

$$y=\dfrac{xp^2}{p+1}$$

Differentiate both sides with respect to $x$

$$y'=\dfrac{(p^2+2xpp')(p+1)-xp^2p'}{(p+1)^2}$$

$$p^3+2p^2+p=p^3+p^2+xpp'(2p+2-p)$$

$$p(p+1)=xpp'(p+2)$$

If $p=0$

$$x(0)^2-y(0)-y=0=>y=0$$

If $p\not=0$

$$p+1=p'(p+2)x$$

$$p'(\dfrac{p+2}{p+1})=\dfrac{1}{x}$$

$$dp(1+\dfrac{1}{p+1})=dx(\dfrac{1}{x})$$

$$p+\ln|p+1|=\ln|x|-\ln C$$

$$x=C(p+1)e^p$$

$$p=\dfrac{y\pm\sqrt{y^2+4xy}}{2x}$$

$$x=C(\dfrac{y\pm\sqrt{y^2+4xy}}{2x}+1)e^{({y\pm\sqrt{y^2+4xy} \over 2x})}$$