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# 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 $$y=e^x$$

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