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).

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