This question is not about solving the ODE, but about understanding something about the Wronskian and Abel's formula.
Given the equation:
$y''-\frac{3}{x}y'+\frac{5}{x^2}y=3$
I get:
$y_1=x^2\cos \left(\ln \left(x\right)\right)$
$y_2=x^2\sin \left(\ln \left(x\right)\right)$
$y_1'=2x\cos \left(\ln \left(x\right)\right)-x\sin \left(\ln \left(x\right)\right)$
$y_2'=2x\sin \left(\ln \left(x\right)\right)+x\cos \left(\ln \left(x\right)\right)$
Computing the Wronskian with denterminant I get:
$W\left(x_1,x_2\right)=\det\begin{pmatrix}\ y_1 & y_2\\ \ y_1' & \ y_2' \end{pmatrix}=x^3$
But computing it with Abel's formula I get:
$W\left(y_1, y_2\right)\left(x\right)=W\left(y_1, y_2\right)\left(1\right)e^{-∫_1^x \frac{-3}{t}dt}\,=|x|^3$
Why there is a difference? Should I omit the absolute value? Am I allowed do that? Why? Abel's formula is supposed to work in any case, isn't it?