This is the problem: (1-x^2 ) y''-2xy'+2y=x-x^3,\qquad y(2)=0,\qquad y(4)=1.
So by inspection I know one of the solutions for the homogenous is $y_1 = x$. Now, to find the other solution I used the following equations:
$ U = \frac{1}{y_1^2}e^{-\int p dx} $
$ y_2 = y_1 \int_{} U dx $
Where $ y_1=x $ and $\displaystyle p= \frac{-2x}{1-x^2} $
And the result I found is:
$ y_2 = 1+ \frac{x}{2} \ln(1-x)-\frac{x}{2} \ln(1+x) $
It works very well... except it will not work for $x>1$ and the initial values I have are $y(2)=0$ and $y(4)=1$. Now, I just checked that this solution works too: $ y_2 = 1+ \frac{x}{2} \ln(x-1)-\frac{x}{2} \ln(1+x) $ and it's actually the solution I need. The problem is I don't know how to get that solution. I mean, I don't want to say "so here you just change $ \displaystyle \frac{x}{2} \ln(1-x)$ for $\displaystyle \frac{x}{2} \ln(x-1) $ and you have it!".
Well, that's all. I know how to do the rest.