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**#28,pg.174:** Find a second solution of
$$(x-1)y''-xy'+y=0$$
using reduction of order and a given solution $y_1(x)=e^x$.

**Solution:** Reduction of order tells us we should guess $y_2(x)=v(x)e^x$. Thus $y_2'(x)=(v'+v)e^x$ and $y_2''(x)=(v''+2v'+v)e^x$. Plus this into the DE to get
$$(x-1)e^x (v''+2v'+v)-xe^x(v'+v)+ve^x=0.$$
Divide by $e^x$ and simplify to get
$$(x-1)v''+(x-2)v'=0.$$
Choose new variable $w=v'$ so that $w'=v''$ and we see
$$(x-1)w'+(x-2)w=0.$$
This is a separable first order differential equation.
To solve it compute
$$\displaystyle\int \dfrac{1}{w} dw = \displaystyle\int \dfrac{2-x}{x-1} dx.$$
The left hand side reduces to $\log |w|$ while the right hand side reduces to $\log |x-1|-x$ (to see it, use $u$-substitution $u=x-1$). This we have
$$\log |w| = \log |x-1|-x+C$$
or
$$|w| = e^C|x-1|e^{-x}$$
or
$$w = \pm e^C (x-1)e^{-x}=A(x-1)e^{-x},$$
for some constant $A$. Thus we recall $w=v'$ and integrate (by parts) to get
$$v = A\displaystyle\int (x-1)e^{-x} dx=-Ae^{-x}x.$$
This suggests our second solution $y_2(x)$ is given by
$$y_2(x)=v(x)e^x=(-Ae^{-x}x+B)e^x=Ax.$$
Thus the general solution is
$$y(x)=c_1e^x+c_2x.$$