General Solution for an ODE I would like to know if someone could provide me the general solution of the following equation:
\begin{align*}
\frac{d^{2}w}{dx^{2}}\cdot\frac{dw}{dx} = e^{-x}w
\end{align*}
Where $w > 0$, $w''>0$ and $x\in[0,1]$. If it is possible, I would also be grateful if someone could solve the next equation:
\begin{align*}
\left(\frac{d^{2}w}{dx^{2}}\right)^{2}\frac{dw}{dx} = e^{-x}w
\end{align*}
Where $w > 0$ and $x\in[0,1]$. Any help is appreciated. Thanks in advance.
 A: Hint:
For $\dfrac{d^2w}{dx^2}\dfrac{dw}{dx}=e^{-x}w$ ,
$\dfrac{1}{w}\dfrac{d^2w}{dx^2}\dfrac{1}{w}\dfrac{dw}{dx}=e^{-x}w^{-1}$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0511.pdf or http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=509.
Let $\begin{cases}u=\dfrac{1}{w}\dfrac{dw}{dx}\\v=e^{-x}w^{-1}\end{cases}$ ,
Then $v(u+1)\dfrac{du}{dv}=u^2-\dfrac{v}{u}$
$(u^3-v)\dfrac{dv}{du}=u(u+1)v$
Let $z=u^3-v$ ,
Then $v=u^3-z$
$\dfrac{dv}{du}=3u^2-\dfrac{dz}{du}$
$\therefore z\left(3u^2-\dfrac{dz}{du}\right)=u(u+1)(u^3-z)$
$3u^2z-z\dfrac{dz}{du}=u^4(u+1)-u(u+1)z$
$z\dfrac{dz}{du}=u(4u+1)z-u^4(u+1)$
This belongs to an Abel equation of the second kind.
For $\left(\dfrac{d^2w}{dx^2}\right)^2\dfrac{dw}{dx}=e^{-x}w$ ,
$\left(\dfrac{1}{w}\dfrac{d^2w}{dx^2}\right)^2\dfrac{1}{w}\dfrac{dw}{dx}=e^{-x}w^{-2}$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0511.pdf or http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=509.
Let $\begin{cases}u=\dfrac{1}{w}\dfrac{dw}{dx}\\v=e^{-x}w^{-2}\end{cases}$ ,
Then $v(2u+1)\dfrac{du}{dv}=u^2\pm\dfrac{\sqrt{v}}{\sqrt{u}}$
