I consider the following ODE $$ f'''(x) = B f''(x) + (Ce^x+D)f'(x) + (Ee^x+F)f(x) $$ for some constants $B$, $C$, $D$, $E$, $F$ and $x\in [0, \infty)$.
This ODE seems to be complicated so I used WolframAlpha (https://www.wolframalpha.com/input/?i=f%27%27%27%28x%29+%3D+B+f%27%27%28x%29+%2B+%28C+e%5Ex+%2B+D%29f%27%28x%29+%2B+%28E+e%5Ex+%2B+F%29+f%28x%29) to find the general form of its solution.
My problem is that I am not sure if the solution that WolframAlpha returns is correct. Can anyone suggest me how can I find the general form of this solution or which software (similar to WolframAlpha) I can use to find this solution.
Let $t=e^x$ ,
Then $\dfrac{df}{dx}=\dfrac{df}{dt}\dfrac{dt}{dx}=e^x\dfrac{df}{dt}=t\dfrac{df}{dt}$
$\dfrac{d^2f}{dx^2}=\dfrac{d}{dx}\left(t\dfrac{df}{dt}\right)=\dfrac{d}{dt}\left(t\dfrac{df}{dt}\right)\dfrac{dt}{dx}=\left(t\dfrac{d^2f}{dt^2}+\dfrac{df}{dt}\right)t=t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}$
$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}\right)=\dfrac{d}{dt}\left(t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}\right)\dfrac{dt}{dx}=\left(t^2\dfrac{d^3f}{dt^3}+3t\dfrac{d^2f}{dt^2}+\dfrac{df}{dt}\right)t=t^3\dfrac{d^3f}{dt^3}+3t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}$
$\therefore t^3\dfrac{d^3f}{dt^3}+3t^2\dfrac{d^2f}{dt^2}+t\dfrac{df}{dt}=Bt^2\dfrac{d^2f}{dt^2}+Bt\dfrac{df}{dt}+(Ct+D)t\dfrac{df}{dt}+(Et+F)f$
$t^3\dfrac{d^3f}{dt^3}+(3-B)t^2\dfrac{d^2f}{dt^2}-(Ct^2+(B+D-1)t)\dfrac{df}{dt}-(Et+F)f=0$
Seek for power series solution