Solve the differential equation $4y=x^2+(y')^2$ 
Solve the differential equation $4y=x^2+(y')^2$.

My try:
$$4 y'=2x+2y'y''$$ hence $2y'=x+y'y''$. Let $y'=vx$, then $y''=v+xv'$, hence
$$2vx=x+vx(v+xv')$$
$$-\int {v\over (v-1)^2} dv=\int {1\over x} dx$$
For R.H.S. Let $u=v-1$
So :
$${1\over u}-\ln u=\ln x+c$$
$${1\over {v-1}}-\ln |v-1|=\ln |x|+c$$
By substitute $v={y'\over x}$ :
$${x\over {y'-x}}-\ln \left|{{y'-x}\over x}\right|=\ln |x| +c$$
that is:
$${x\over {y'-x}}-\ln \left|y'-x\right|=c$$
True ?

$4y=x^2+(y')^2$
$\left(\dfrac{dy}{dx}\right)^2=4y-x^2$
$\dfrac{dy}{dx}=\pm\sqrt{4y-x^2}$
Let $u=\pm\sqrt{4y-x^2}$ ,
Then $y=\dfrac{u^2+x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}=u$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{2u-x}{2}$
$\dfrac{du}{dx}=2-\dfrac{x}{u}$
Let $v=\dfrac{u}{x}$ ,
Then $u=xv$
$\dfrac{du}{dx}=x\dfrac{dv}{dx}+v$
$\therefore x\dfrac{dv}{dx}+v=2-\dfrac{1}{v}$
$x\dfrac{dv}{dx}=-v+2-\dfrac{1}{v}$
$x\dfrac{dv}{dx}=-\dfrac{v^2-2v+1}{v}$
$x\dfrac{dv}{dx}=-\dfrac{(v-1)^2}{v}$
$\dfrac{v}{(v-1)^2}~dv=-\dfrac{dx}{x}$
$\int\dfrac{v}{(v-1)^2}~dv=-\int\dfrac{dx}{x}$
$\ln(v-1)-\dfrac{1}{v-1}=-\ln x+c$
$(v-1)e^{-\frac{1}{v-1}}=\dfrac{C_1}{x}$
$\left(\dfrac{u}{x}-1\right)e^{-\frac{1}{\frac{u}{x}-1}}=\dfrac{C_1}{x}$
$(u-x)e^\frac{x}{x-u}=C_1$
$(\pm\sqrt{4y-x^2}-x)e^\frac{x}{x\mp\sqrt{4y-x^2}}=C_1$
$(x\pm\sqrt{4y-x^2})e^\frac{x}{x\pm\sqrt{4y-x^2}}=C$
