How to solve this differential equation $y(t) = y′(t)+ \frac{e^{2t}}{y'(t)}$ I don't understand what a type of this equation and which method I could use for:
$$y(t) = y′(t)+ \frac{e^{2t}}{y'(t)}$$
Please, help me.
 A: Introduce $x:=e^t$ as new independent variable. Then
$$y'={dy\over dt}={dy\over dx}\,{dx\over dt}=x\dot y\ ,$$
where the dot denotes differentiation with respect to $x$. In this way we obtain the new differential equation
$$y=x\dot y+{x^2\over x\dot y}=x\left(\dot y+{1\over\dot y}\right)\ .$$
Now solve for $\dot y$ and get
$$\dot y={1\over2}\left({y\over x}\pm\sqrt{\left({y\over x}\right)^2-4}\right)\ .$$
This is a so-called homogeneous ODE. Writing $y:=u\,x$ with a new unknown function $u$ makes it separable.
A: Solving
$$
y'^2-yy'+e^{2t}=0\tag{1}
$$
for $y'$ gives
$$
y'=\frac{y\pm\sqrt{y^2-4e^{2t}}}2\tag{2}
$$
This suggests the substitution $y=ue^t$, which gives
$$
u'=\frac{-u\pm\sqrt{u^2-4}}2\tag{3}
$$
and this can be solved by
$$
t-t_0=\log\left(u\pm\sqrt{u^2-4}\right)-\frac u4\left(u\pm\sqrt{u^2-4}\right)\tag{4}
$$
Back substitution gives a plain equation, but the expression of $y$ in terms of $t$ is still not simple.

Derivation of $\boldsymbol{(4)}$
For the "$+$" version of $(4)$, use $u=2\sec(v)$ and $s=\sin(v)$ so that $s=\frac{\sqrt{u^2-4}}u$ :
$$
\begin{align}
t
&=\int\frac{2\,\mathrm{d}u}{-u+\sqrt{u^2-4}}\\
&=\int\frac{2\tan(v)\sec(v)\,\mathrm{d}v}{-\sec(v)+\tan(v)}\\
&=-2\int\frac{\sin(v)\,\mathrm{d}\sin(v)}{(1-\sin^2(v))(1-\sin(v))}\\
&=-2\int\frac{s\,\mathrm{d}s}{(1+s)(1-s)^2}\\
&=\frac12\int\left(\frac1{1-s}+\frac1{1+s}-\frac2{(1-s)^2}\right)\mathrm{d}s\\
&=\frac12\log\left(\frac{1+s}{1-s}\right)-\frac1{1-s}+C\\[1pt]
&=\frac12\log\left(\frac{u+\sqrt{u^2-4}}{u-\sqrt{u^2-4}}\right)-\frac u{u-\sqrt{u^2-4}}+C\\[2pt]
&=\log\left(u+\sqrt{u^2-4}\right)-\frac u4\left(u+\sqrt{u^2-4}\right)+C-\log(2)
\end{align}
$$
The derivation for the "$-$" version of $(4)$ similar
