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I'm trying to solve this 3rd order non h ode.

$$ x^3y'''-3x^2y''+6xy'-6y=2x^4e^x $$

But im stuck after solving the homogeneous

$ x^3y'''-3x^2y''+6xy'-6y=0 $

With $y=e^a$ for $a=kx$

My output was $k^3x^3-k^2 3x^2+k6x-6=0$

But Im not sure how to follow up since I dont usually have functions of x before the differentials.

EDIT: Ok I solved the homo transforming $y(x) = u(lnx)$

result was $y(x)=ax+bx^2+cx^3$.

What method should I use for getting the particular solution?

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    This is an Euler type ODE. The relevant solution class is $y=x^\lambda$. Or set $x=e^t$ to transform the ODE into a standard linear ODE with constant coefficients. – 2017-01-14
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    Show that $$x,x^2,x^3$$ are solutions of the homogenious part like the ansatz written above – 2017-01-14
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    DO you mean replace x with $$e^t$$ at the very beginning?? @LutzL – 2017-01-14
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    Yes. Of course everywhere. Which means to substitute $u(t)=y(e^t)$ or equivalently $y(x)=u(\ln(x))$ and compute the derivatives via chain rule.. – 2017-01-15

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