Made a slip somewhere in the tutorial and rather looking for it I said I’d put it up here — the point is that the $u(x)$ term disappears so we have a differential equation in $u'(x)$ and $u''(x)$ — in other words a function and it’s derivative. This can then be integrated in the usual way.

Verify that $y_1(x)=x-1$ is a solution to the second order ODE

$(x-1)^2\frac{d^2y}{dx^2}-(x^2-1)\frac{dy}{dx}+(x+1)y=0.$  (*)

Solution

We have that $y'_1(x)=1$ and $y''_1(x)=0$. Hence

$(x-1)^2y''_1(x)-(x^2-1)y'_1(x)+(x+1)y_1(x)$

$=-(x^2-1)+(x+1)(x-1)=0$

as required. That is $y_1(x)$ is a solution.

Find the general solution by trying a solution of the form $y_2(x)=(x-1)u(x)$

Let $y_2(x)=(x-1)u(x)$. Hence

$y'_2(x)=(x-1)u'(x)+u(x)$.

$y''_2(x)=(x-1)u''(x)+2u'(x).$

Substituting into the left-hand side of (*):

$(x-1)^2\left[(x-1)u''(x)+2u'(x)\right]-(x^2-1)\left[(x-1)u'(x)+u(x)\right]+(x+1)(x-1)u(x)$

Grouping into $u''(x),u'(x)$ and $u(x)$:

$u''(x)\left[(x-1)^4\right]+u'(x)\left[2(x-1)^2-(x^2-1)(x-1)\right]+$

$u(x)\underbrace{\left[-(x^2-1)+(x+1)(x-1)\right]}_{=0}$

A bit of factorisation:

$u''(x)(x-1)^3-(x-1)^3u'(x)\overset{?}{=}0$

$\Rightarrow u''(x)-u'(x)=0$

Now let $v(x)=u'(x)$ and note $u''(x)=v'(x)$:

$v'(x)-v(x)=0$

$\Rightarrow \frac{dv}{dx}=v$

$\Rightarrow \frac{dv}{v}=dx$

$\Rightarrow \int \frac{dv}{v}=\int dx$

$\Rightarrow \ln|v|=x+C$.

$\Rightarrow e^{x+C}=e^xe^C=C_1e^x=v$,

That is

$v=u'(x)=\frac{du}{dx}=C_1e^{x}$

$\Rightarrow du=C_1e^x\,dx$

$\Rightarrow \int du=C_1\int e^x\,dx$

$\Rightarrow u(x)=C_1e^{x}+C_2$,

for a general solution:

$y_g(x)=(x-1)(C_1e^x+C_2)$.