Solve the IVP $ y^{\prime \prime}=y^{\prime} e^y$, $y(0)=0$, $y^{\prime}(0)=1$ The solution $y(x)$ of the initial value problem
$$
y^{\prime \prime}=y^{\prime} e^y, \quad y(0)=0, y^{\prime}(0)=1,
$$
is $y(x)=...$
How to solve this? I stuck at $\int \frac{y''}{y'}dt=\int e^y dt+C$
$\Rightarrow \ln y'(t)=\int e^y dt+C$
$\Rightarrow y'(t)=Ae^{\int e^y dt}$
$\Rightarrow y(t)=A\int\int e^{\int e^y dt}+c$.
 A: I assume a connected domain.
By inspection, we know: $$y'=e^y+A$$For some constant $A$.
Now this is harder. Consider: $$-y'e^{-y}=-1-Ae^{-y}$$Substitute $v=e^{-y}$ - we want to solve: $$v'=-1-Av$$Which is a standard first order linear ODE, we can use the method of integrating factors. $$v'+Av=-1\implies(v\cdot e^{At})'=-e^{At}$$And: $$v=\begin{cases}Be^{-At}-\frac{1}{A}&A\neq0\\B-t&A=0\end{cases}$$For some constant $B$. Finally: $$y=-\ln v$$
From $y(0)=0$, we know $v(0)=\pm1$. From $y'(0)=1$, we know $-1=-\frac{v'(0)}{v}=\mp v'(0)$. In the case $A=0$, that would mean that $1=\pm(-1)$, so we take the minus sign and $v(0)=+1$, so that $B=1$. If $A\neq0$, we know $B-1/A=\pm1$ and $-AB=-1$. We get $AB-1=\pm A\therefore0=\pm A,A=0$, which is a contradiction. For these initial conditions, this case never arises.
Therefore I claim: $$y(t)=-\ln(1-t)$$
Defined on $t\in(-\infty,1)$, is the only solution to your problem.
A: Reduction of order with $v:=y'\implies y'=vv'$. Thus, the ODE can be written as $e^y v=vv' \iff v(v'-e^y)=0$. Thus, $v\equiv 0$ or $v'=e^y$.
Casework:

*

*If $v\equiv 0$, then $y\equiv c_1$ for some arbitrary constant $c_1\not=0$. Since $y'(0)=1$, then it is not a solution for the IVP.

*If $v'=e^y$, then $v=e^y +c_2$ for some arbitrary constant $c_2$. Thus $y'=e^y +c_2$ but since $y'(0)=1$ and $y(0)=0$, then $1=e^0+c_2\iff c_2=0$. Thus $y'=e^y\implies \int \frac{1}{e^y}\, dy=\int 1\, dx \iff -e^{-y}=x+c_3 $ but $y(0)=0$, then $-1=c_3$. Hence $-e^{-y}=x-1\iff e^{-y}=1-x\iff -y=\log(1-x)$. Thus, $y(x)=-\log(1-x)$ for $1-x>0$.

