How find this equation $y''+(y')^2\cdot e^x=0$ if the ODE 
$$y''+(y')^2\cdot e^x=0$$ such $$y(0)=1,y'(0)=1$$
Find the $y(x)=?$
my  ugly methods: let $$y'=p,y''(x)=p'(x)$$
so
$$p'(x)+p^2\cdot e^x=0$$
$$\dfrac{dp}{dx}=-p^2e^x$$
$$\dfrac{dp}{-p^2}=e^xdx$$
$$\Longrightarrow \dfrac{1}{p}=e^x+c\Longrightarrow p=e^{-x}(p(0)=1)$$
so
$$y(x)=2-e^{-x}$$
My question: This ODE have other methods,? such I want use
$$(y'\cdot e^x)'=y''e^x+e^xy'$$
Thank you 
 A: Rewrite:
$$
\begin{align}
y''+(y')^2e^x&=0\\
y''&=-(y')^2e^x\\
\frac{d(y')}{dx}&=-(y')^2e^x\\
-\frac{d(y')}{(y')^2}&=e^x\ dx\\
-\int\frac{d(y')}{(y')^2}&=\int e^x\ dx\\
\frac{1}{y'}&=e^x+C_1\\
y'&=\frac{1}{e^x+C_1}\\
\frac{dy}{dx}&=\frac{1}{e^x+C_1}\\
\int\  dy&=\int \frac{1}{e^x+C_1}\ dx\\
&=\int \frac{e^{-x}}{1+C_1e^{-x}}\ dx.
\end{align}
$$
Now, let $u=1+C_1e^{-x}$ and $du=-C_1e^{-x}\ dx$, then
\begin{align}
\int\  dy&=\int \frac{e^{-x}}{1+C_1e^{-x}}\ dx\\
\int\  dy&=\int \frac{e^{-x}}{u}\cdot \frac{du}{-C_1e^{-x}}\\
\int\  dy&=-\frac{1}{C_1}\int \frac{du}{u}\\
y&=-\frac{1}{C_1}\ln (1+C_1e^{-x})+C_2\\
y&=\frac{x-\ln(e^x+C_1)}{C_1}+C_2.
\end{align}
A: Rewrite:
$$
\begin{align}
y''+(y')^2e^x&=0\\
y''&=-(y')^2e^x\\
\frac{d(y')}{dx}&=-(y')^2e^x\\
-\frac{d(y')}{(y')^2}&=e^x\ dx\\
-\int\frac{d(y')}{(y')^2}&=\int e^x\ dx\\
\frac{1}{y'}&=e^x+C_1\\
y'(x)&=\frac{1}{e^x+C_1}\\
\end{align}
$$
For $y'(0)=1$, we get $C_1=0$ and $y'(x)=e^{-x}$. Hence
\begin{align}
\frac{dy}{dx}&=e^{-x}\\
\int\  dy&=\int e^{-x}\ dx\\
y(x)&=-e^{-x}+C_2
\end{align}
For $y(0)=1$, we get $C_2=2$ and $y(x)=2-e^{-x}$.

Credit answer : Mr. Tunk-Fey (>‿◠)✌
A: My first thought when I saw your questions was that it is non-linear differentiation and there's an $e^x$ term; as such, I would try
$$y(x) = e^{k x}$$
where $k$ is such that (by considering the powers/primes in the ODE)
$$k + 2k + 1 = 0 \Rightarrow k = -1.$$
This then gives the same result as yours (after using the initial condition).
Also a good idea would be to let, for example, $Y = y'$, and solve the 1st order ODE for $Y$, then integrate for $y'$.
Hope this helps! If so, please remember to upvote and/or accept this answer! =D
