Third Order Differential Equation (Change of variable) I'm trying to find a suitable change of variable for 
$yy'''+3y'y''=2e^x$. But no idea, I tried almost everything... e.g. $x=\log t$   but the problem is getting huge.
Any ideas?
Thanks
 A: First of all, notice that:
\begin{align*}
&yy^{\prime\prime\prime} + 3y^{\prime}y^{\prime\prime} = 2e^{x} \Leftrightarrow \int yy^{\prime\prime\prime}\mathrm{d}x + 3\int y^{\prime}y^{\prime\prime}\mathrm{d}x = 2\int e^{x}\mathrm{d}x \Leftrightarrow\\
& yy^{\prime\prime} - \int y^{\prime}y^{\prime\prime}\mathrm{d}x + 3\int y^{\prime}y^{\prime\prime}\mathrm{d}x = 2e^{x} \Leftrightarrow yy^{\prime\prime} + (y^{\prime})^{2} = 2e^{x} + k
\end{align*}
If we repeat the process, we get:
\begin{align*}
&\int yy^{\prime\prime}\mathrm{d}x + \int y^{\prime}y^{\prime}\mathrm{d}x = 2\int e^{x}\mathrm{d}x + \int k\mathrm{d}x \Leftrightarrow\\
& yy^{\prime} - \int y^{\prime}y^{\prime}\mathrm{d}x + \int y^{\prime}y^{\prime}\mathrm{d}x = 2e^{x} + kx \Leftrightarrow yy^{\prime} = 2e^{x} + kx + c \Leftrightarrow\\
&\int yy^{\prime}\mathrm{d}x = 2e^{x} + \frac{kx^{2}}{2} + cx + d \Leftrightarrow  (y)^{2} = 4e^{x} + kx^{2} + 2cx + 2d \Leftrightarrow\\
& y(x) = \pm\sqrt{4e^{x} + kx^{2} + 2cx + 2d}
\end{align*}
A: $$yy'''+3y'y''=2e^x$$
$$yy'''+y'y''+2y'y''=2e^x$$
$$(yy'')'+2y'y''=2e^x$$
$$(yy'')'+(y'^2)'=2e^x$$
Integrate
$$yy''+y'^2=2e^x+K_1$$
$$(y'y)'=2e^x+K_1$$
Integrate again
$$y'y=2e^x+K_1x+K_2$$
$$(y^2)'=4e^x+C_1x+C_2$$
Integrate again....
$$y^2=4e^x+R_1x^2+R_2x+R_3$$
$$\boxed {y(x)=\pm \sqrt {4e^x+R_1x^2+R_2x+R_3}}$$
