<s>[INST] variation of parameters for a Cauchy-Euler problem: where am I wrong? The problem is this one
$$x^2y'' - xy' -3y = 5x^4.$$
Then the complementary solution I get is 
$$c_1x + c_2x\ln x.$$
Then after that the particular solution is
$$\ln x e^{4x} \left(\frac{x}{4} - \frac{1}{16}\right) - \frac{e^{4x}}{16} - \frac{1}{16} \int \frac{e^{4x}}{x}dx.$$
My doubt is how can integrate the last part or where am I doing it wrong
 [/INST] A.N.Other method is
$$
x^2y'' - xy' - 3y = 5x^4
$$
we can realise thath
$$
\dfrac{d}{dx}x^2y' = x^2y'' + 2xy'\implies x^2y'' = \dfrac{d}{dx}x^2y'  - 2xy'
$$
so we can insert into the original equation to yeild
$$
\dfrac{d}{dx}x^2y'  - 2xy' - xy' -3y = \dfrac{d}{dx}\left(x^2y'\right) - 3(xy' + y) = 5x^4\\
 \dfrac{d}{dx}\left(x^2y'\right) -\dfrac{d}{dx}(xy) = 5x^4 
$$
or 
$$
y' -\frac{3}{x}y = x^3 + \frac{C_1}{x^2}
$$
integrating factor
$$
y\mathrm{e}^{-3\ln x} = \int x^3 \mathrm{e}^{-3\ln x} dx +C_1 \int \frac{1}{x^2}\mathrm{e}^{-3\ln x} dx +C_2\\
y\frac{1}{x^3} = \int x^3\frac{1}{x^3} dx + C_1 \int \frac{1}{x^2}\frac{1}{x^3}dx + C_2\\
\frac{y}{x^3} = \int 1 dx +C_1 \int x^{-5} + C_2 = \frac{1}{2}x^2 -\frac{C_1}{4}x^{-4} + C_2
$$
thus
$$
y = x^4 + C_2x^3 -\frac{C_1}{4}x^{-1} 
$$
As obtained in the previous solutions.
</s>