third order differential equation Could any experts give me some suggestions on solving this 3 order differential equation?                                                                      $$x'''-x''-(x')(e^{2t})-(x)(e^{2t})=0$$
 A: You can reduce it a little by the trick 
$x'''-x''=(x+x')e^{2t}\iff e^{-t}(x'''-x'')=e^t(x'+x)\implies \left(e^{-t}x''\right)'=\left(e^{t}x\right)'$
So you get $x''=e^{2t}x+ce^t$
But I'm afraid that what we got now is not very attractive either...
A: $$x'''-x''-e^{2t}x'-e^{2t}x=0 \tag 1$$
Change of variable : $\quad z=e^t \quad;\quad \frac{dx}{dt}=z\frac{dx}{dz}\quad;\quad \frac{d^2x}{dt^2}=z^2\frac{d^2x}{dz^2}+z\frac{dx}{dz}\quad;\quad \frac{d^3x}{dt^3}=z^3\frac{d^3x}{dz^3}+3z^2\frac{d^2x}{dz^2}+z\frac{dx}{dz}$
$$\left(z^3\frac{d^3x}{dz^3}+3z^2\frac{d^2x}{dz^2}+z\frac{dx}{dz}\right)-\left(z^2\frac{d^2x}{dz^2}+z\frac{dx}{dz}\right)-z^2\left(z\frac{dx}{dz}\right)-z^2x=0$$
After simplification :
$$z\frac{d^3x}{dz^3}+2\frac{d^2x}{dz^2}-z\frac{dx}{dz}-x=0$$
$$\frac{d}{dz}\left(z\frac{d^2x}{dz^2}+\frac{dx}{dz}-zx\right)=0$$
$$z\frac{d^2x}{dz^2}+\frac{dx}{dz}-zx=c_1$$
$$\frac{d^2x}{dz^2}+\frac1z\frac{dx}{dz}-x=\frac{c_1}{z} \tag 2$$
The homogeneous part is a Bessel ODE which solution is :
$$x_h=c_2J_0(iz)+c_3Y_0(-iz)$$
$J_0$ and $Y_0$ are the Bessel functions of order $0$ and first and second kind.
$$x(z)=c_2J_0(iz)+c_3Y_0(-iz)+c_1x_p(z)$$
The general solution of Eq.$(1)$ is on the form :
$$x(t)=c_2J_0(ie^t)+c_3Y_0(-ie^t)+c_1x_p(e^t)$$
$x_p(z)$ is any particular solution of Eq.$(3)$.
$$\frac{d^2x_p}{dz^2}+\frac1z\frac{dx_p}{dz}-x_p=\frac{1}{z}\tag 3$$
Theoretically it is possible to find $x_p(z)$ thanks to the method of variation of parameters. Let $x_p(z)=f(z)J_0(iz)+g(z)Y_0(-iz)$ and determine $f(z)$ and $g(z)$ so that the reduction of order occurs. This is an awfully arduous task, but possible anyways. Sorry I am too lazy to do it. I let my  faithful slave do the job.
Thanks to WolframAlpha :
$$x_p(z)=\frac{\pi^2}{4}zH_0(iz)J_0(iz)Y_1(-iz)+\frac{\pi^2}{4}zH_{-1}(iz)J_0(iz)Y_0(-iz) +\frac{\pi}{2}zY_0(-iz)\:_2F_1\left(\frac12;1,\frac{3}{2};\frac{z^2}{4} \right)$$
$H_0$ and $H_{-1}$ are Struve functions.
It is probably possible to simplify this horrifying result. But my faithful slave seems so tired that he refuses further calculus.
