Solve second order differential equations There are two differential equations that I could not solve. Can someone please help me solve them?
$$
(x^2+y^2)y′′-y(y^{′})^3+xy′-y=0   
$$
and
$$
xy^2y′′+2y^2y′-4xy(y^{′})^2+2x^2(y^{′})^3=0,
$$
where $y=y(x)$.
Thanks in advance!
 A: $$xy^2y′′+2y^2y′-4xy(y')^2+2x^2(y')^3=0$$
Obviously $y=x$ is a particular solution. So, let $y(x)=x\:z(x)$
$y'=z+xz'$ and $y''=2z'+xz''$ that we put into the ODE. After simplification :
$$xz^2z''+2x^2(z')^3+2xz(z')^2+2z^2z'=0$$
Since the ODE is homogeneous with respect to $z, z',z''$, let $z(x)=e^{g(x)}$
$z'=e^g g'$ and $z''=e^g (g''+g'^2)$ that we put into the ODE. After simplification :
$$xg''+2x^2(g')^3+3x(g')^2+2g'=0$$
The ODE is reduced to a first order ODE with $G(x)=g'$
$$xG'+2x^2G^3+3xG^2+2G=0$$
We observe that the ODE is homogeneous in the particular case of the form $G=\frac{a}{x}$. Looking for a particular solution, we determine $a=-\frac{1}{2}$. So a particular solution is $G=-\frac{1}{2x}$. This draw us to the change of function:
$G(x)=H(x)-\frac{1}{2x}$ that we put into the ODE. After simplification :
$$2xH'+4x^2H^3+H=0$$
The new ODE is of the Bernoulli kind. Thanks to the classical method of solving the Bernoulli's equations, this leads to :
$$H=\frac{1}{\sqrt{4x^2+cx}}$$
$$G=H-\frac{1}{2x}=\frac{1}{\sqrt{4x^2+cx}}-\frac{1}{2x}=\frac{dg}{dx}$$
After integration :
$$g=\frac{1}{2}\ln(4\sqrt{4x^2+cx}+8x+c)-\frac{1}{2}\ln(x)+constant$$
$$z=e^g=C \sqrt{\frac{4\sqrt{4x^2+cx}+8x+c}{x}}$$
$$y=xz=C \sqrt{4x\sqrt{x(4x+c)}+8x^2+cx}=C\sqrt{\left(\sqrt{x(4x+c)}+2x\right)^2}$$
$$y=C\left(\sqrt{x(4x+c)}+2x\right)$$
or, with constants $a,b$ :
$$y=a\left(\sqrt{x(x+b)}+x\right)$$
A: Hint:
For $(x^2+y^2)y''-y(y')^3+xy'-y=0$ ,
Let $x=e^t$ ,
Then $t=\ln x$
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{dy}{dt}=e^{-t}\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(e^{-t}\dfrac{d^2y}{dt^2}-e^{-t}\dfrac{dy}{dt}\right)e^{-t}=e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}$
$\therefore(e^{2t}+y^2)\left(e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}\right)-y\left(e^{-t}\dfrac{dy}{dt}\right)^3+\dfrac{dy}{dt}-y=0$
$(1+e^{-2t}y^2)\dfrac{d^2y}{dt^2}-(1+e^{-2t}y^2)\dfrac{dy}{dt}-e^{-3t}y\left(\dfrac{dy}{dt}\right)^3+\dfrac{dy}{dt}-y=0$
$(e^{-2t}y^2+1)\dfrac{d^2y}{dt^2}-e^{-2t}y^2\dfrac{dy}{dt}-e^{-3t}y\left(\dfrac{dy}{dt}\right)^3-y=0$
Let $y=e^tu$ ,
Then $\dfrac{dy}{dt}=e^t\dfrac{du}{dt}+e^tu$
$\dfrac{d^2y}{dt^2}=e^t\dfrac{d^2u}{dt^2}+e^t\dfrac{du}{dt}+e^t\dfrac{du}{dt}+e^tu=e^t\dfrac{d^2u}{dt^2}+2e^t\dfrac{du}{dt}+e^tu$
$\therefore(u^2+1)\left(e^t\dfrac{d^2u}{dt^2}+2e^t\dfrac{du}{dt}+e^tu\right)-u^2\left(e^t\dfrac{du}{dt}+e^tu\right)-e^{-2t}u\left(e^t\dfrac{du}{dt}+e^tu\right)^3-e^tu=0$
$e^t(u^2+1)\left(\dfrac{d^2u}{dt^2}+2\dfrac{du}{dt}+u\right)-e^tu^2\left(\dfrac{du}{dt}+u\right)-e^tu\left(\dfrac{du}{dt}+u\right)^3-e^tu=0$
$(u^2+1)\left(\dfrac{d^2u}{dt^2}+2\dfrac{du}{dt}+u\right)-u^2\left(\dfrac{du}{dt}+u\right)-u\left(\dfrac{du}{dt}+u\right)^3-u=0$
$(u^2+1)\dfrac{d^2u}{dt^2}+(u^2+2)\dfrac{du}{dt}-u\left(\dfrac{du}{dt}+u\right)^3=0$
Let $v=\dfrac{du}{dt}$ ,
Then $\dfrac{d^2u}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{du}\dfrac{du}{dt}=v\dfrac{dv}{du}$
$\therefore(u^2+1)v\dfrac{dv}{du}+(u^2+2)v-u(v+u)^3=0$
