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 (x-1)y''-xy'+y=0 Solve using the reduction method to order

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

Answers (1)

2020-12-03
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 \(y=e^x\)
0

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