How do I solve $(y')^2-yy'=e^x$

Question

I tried taking the derivative wrt x but it did not help. I searched other methods but I could not find a similar question to this. Thanks in advance.

How to solve the equation :

$$(y')^2-yy'=e^x$$

Answer 1

We can factor the DE as $y'(y-y')=e^x$, and then using the integrating factor $e^{-x}$ we can "factor" the DE further as $y'(e^{-x}y)'=1$. There is an obvious asymmetry between the terms being multiplied, which can be made symmetric using the substitution $y:=e^{x/2}v$; if you substitute that into the DE, use product rule, difference of squares, and cancel $e^{\pm x/2}$s, you can get a nice separable equation in $v$.

Answer 2

Here is how I solved it:

$$ \begin{align} \\ 0&=\frac{d}{dx}\left[\frac{dy}{dx}\left(\frac{dy}{dx}-y\right)\right]-e^{x}\\ \\ &=\frac{d}{dx}\left[\frac{dy}{dx}\left(\frac{dy}{dx}-y\right)\right]-\frac{dy}{dx}\left(\frac{dy}{dx}-y\right)\\ \\ &=\frac{d^{2}y}{dx^{2}}\left(\frac{dy}{dx}-y\right)+\frac{dy}{dx}\left(\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\right)-\frac{dy}{dx}\left(\frac{dy}{dx}-y\right)\\ \\ &=\left(2\frac{dy}{dx}-y\right)\left(\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\right)\\ \\ \end{align} $$

We get two ODEs:

$$ \begin{align} \\ 0&=\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\\ \\ 0&=2\frac{dy}{dx}-y\\ \\ \end{align} $$

These are normal ODEs and the solutions (after substitution to the original equation) are:

$$ \begin{align} \\ y&=Ce^{x}-\frac{1}{C}\\ \\ y&=\pm 2i\cdot e^{x/2}\\ \\ \end{align} $$