# The function y=x is one the solution of (x-1)y''-xy'+y=0 Solve using the reduction method to order

The function y=x is one the solution of
$\left(x-1\right){y}^{″}-x{y}^{\prime }+y=0$
Solve using the reduction method to order
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FieniChoonin
Given y=x is one solution of $\left(x-1\right){y}^{″}-x{y}^{\prime }+y=0$
Let, y=vx be another solution of the given equation.Then
$\left(x-1\right)\left(vx{\right)}^{″}-x\left(vx{\right)}^{\prime }+vx=0$
$\left(x-1\right)\left({v}^{″}+2{v}^{\prime }\right)-x\left(v+x{v}^{\prime }\right)+vx=0$
$\left(x-1\right)x{v}^{″}+2\left(x-1\right){v}^{\prime }-{x}^{2}{v}^{\prime }=0$
$\left(x-1\right)x{v}^{″}+\left(2x-2-{x}^{2}\right){v}^{\prime }=0$
Now suppose ${v}^{\prime }=u$. On substituting
$\left(x-1\right)x{u}^{\prime }+\left(2x-2-{x}^{2}\right)u=0$
$\frac{du}{u}=-\frac{2x-2-{x}^{2}}{\left(x-1\right)x}dx$
$\frac{du}{u}=\frac{{x}^{2}-2x+2}{\left(x-1\right)x}dx$
$\frac{du}{u}=\frac{\left(x-1{\right)}^{2}+1}{\left(x-1\right)x}dx$
$\frac{du}{u}=\frac{x-1}{x}-\left(\frac{1}{x}-\frac{1}{x-1}\right)dx$
$\int \frac{du}{u}=\int \left(1-\frac{1}{x}\right)dx-\int \frac{dx}{x}+\int \frac{dx}{x-1}$
$\mathrm{ln}u=x-2\mathrm{ln}x+\mathrm{ln}\left(x-1\right)+\mathrm{ln}c$ [c=integrating constant]
$\mathrm{ln}u=x+\mathrm{ln}\frac{c\left(x-1\right)}{{x}^{2}}$
$u=\frac{c\left(x-1\right)}{{x}^{2}{e}^{x}}$
$\frac{dv}{dx}=c\left[\frac{1}{x}-\frac{1}{{x}^{2}}\right]{e}^{x}$
$dv=c\left[\frac{1}{x}-\frac{1}{{x}^{2}}\right]{e}^{x}dx$
$\int dv=c\int \left[\frac{1}{x}-\frac{1}{{x}^{2}}\right]{e}^{x}dx$
$v=\frac{c{e}^{x}}{x}+d$ [$\int \left[f\left(x\right)+{f}^{\prime }\left(x\right)\right]{e}^{x}dx={e}^{x}f\left(x\right)+c$], [d=integrating constant]
Therefore
$y=vx$
$=\left(\frac{c{e}^{x}}{x}+d\right)x$
$=c{e}^{x}+dx$
Therefore, another linearly independent solution is $y={e}^{x}$