Given y=x is one solution of \((x-1)y''-xy'+y=0\)

Let, y=vx be another solution of the given equation.Then

\((x-1)(vx)''-x(vx)'+vx=0\)

\((x-1)(v''+2v')-x(v+xv')+vx=0\)

\((x-1)xv''+2(x-1)v'-x^2v'=0\)

\((x-1)xv''+(2x-2-x^2)v'=0\)

Now suppose \(v'=u\). On substituting

\((x-1)xu'+(2x-2-x^2)u=0\)

\(\frac{du}{u}=-\frac{2x-2-x^2}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{x^2-2x+2}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{(x-1)^2+1}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{x-1}{x}-(\frac{1}{x}-\frac{1}{x-1})dx\)

\(\int\frac{du}{u}=\int(1-\frac{1}{x})dx-\int\frac{dx}{x}+\int\frac{dx}{x-1}\)

\(\ln u=x-2\ln x+\ln(x-1)+\ln c\) [c=integrating constant]

\(\ln u=x+\ln\frac{c(x-1)}{x^2}\)

\(u=\frac{c(x-1)}{x^2e^x}\)

\(\frac{dv}{dx}=c[\frac{1}{x}-\frac{1}{x^2}]e^x\)

\(dv=c[\frac{1}{x}-\frac{1}{x^2}]e^xdx\)

\(\int dv=c\int[\frac{1}{x}-\frac{1}{x^2}]e^xdx\)

\(v=\frac{ce^x}{x}+d\) [\(\int[f(x)+f'(x)]e^xdx=e^xf(x)+c\)], [d=integrating constant]

Therefore

\(y=vx\)

\(=(\frac{ce^x}{x}+d)x\)

\(=ce^x+dx\)

Therefore, another linearly independent solution is \(y=e^x\)

Let, y=vx be another solution of the given equation.Then

\((x-1)(vx)''-x(vx)'+vx=0\)

\((x-1)(v''+2v')-x(v+xv')+vx=0\)

\((x-1)xv''+2(x-1)v'-x^2v'=0\)

\((x-1)xv''+(2x-2-x^2)v'=0\)

Now suppose \(v'=u\). On substituting

\((x-1)xu'+(2x-2-x^2)u=0\)

\(\frac{du}{u}=-\frac{2x-2-x^2}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{x^2-2x+2}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{(x-1)^2+1}{(x-1)x}dx\)

\(\frac{du}{u}=\frac{x-1}{x}-(\frac{1}{x}-\frac{1}{x-1})dx\)

\(\int\frac{du}{u}=\int(1-\frac{1}{x})dx-\int\frac{dx}{x}+\int\frac{dx}{x-1}\)

\(\ln u=x-2\ln x+\ln(x-1)+\ln c\) [c=integrating constant]

\(\ln u=x+\ln\frac{c(x-1)}{x^2}\)

\(u=\frac{c(x-1)}{x^2e^x}\)

\(\frac{dv}{dx}=c[\frac{1}{x}-\frac{1}{x^2}]e^x\)

\(dv=c[\frac{1}{x}-\frac{1}{x^2}]e^xdx\)

\(\int dv=c\int[\frac{1}{x}-\frac{1}{x^2}]e^xdx\)

\(v=\frac{ce^x}{x}+d\) [\(\int[f(x)+f'(x)]e^xdx=e^xf(x)+c\)], [d=integrating constant]

Therefore

\(y=vx\)

\(=(\frac{ce^x}{x}+d)x\)

\(=ce^x+dx\)

Therefore, another linearly independent solution is \(y=e^x\)