# 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.

• you can also use a numerical method – Dr. Sonnhard Graubner May 31 '17 at 16:32
• Do you know that a general solution exists in closed form for either equation? My gut reaction is that no closed-form solution exists in either case. – Michael Seifert May 31 '17 at 17:01
• Ignoring the conditions, it can be seen that $w=-e^{-x}$ is a solution of the first equation. Perhaps this fact can be used to find a solution that does satisfy the conditions. – Paul May 31 '17 at 19:39

Hint:

For $\dfrac{d^2w}{dx^2}\dfrac{dw}{dx}=e^{-x}w$ ,

$\dfrac{1}{w}\dfrac{d^2w}{dx^2}\dfrac{1}{w}\dfrac{dw}{dx}=e^{-x}w^{-1}$

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 $\left(\dfrac{d^2w}{dx^2}\right)^2\dfrac{dw}{dx}=e^{-x}w$ ,

$\left(\dfrac{1}{w}\dfrac{d^2w}{dx^2}\right)^2\dfrac{1}{w}\dfrac{dw}{dx}=e^{-x}w^{-2}$

Let $\begin{cases}u=\dfrac{1}{w}\dfrac{dw}{dx}\\v=e^{-x}w^{-2}\end{cases}$ ,

Then $v(2u+1)\dfrac{du}{dv}=u^2\pm\dfrac{\sqrt{v}}{\sqrt{u}}$