What is the solution for y^{''}-(x/(x-1))y^'+1/(x-1)y=0 ?

The solution for y^{''}-(x/(x-1))y^'+1/(x-1)y=0 is y=e^{x+c_{1}-1}+c_{2}x

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y^{''}-(x/(x-1))y^'+1/(x-1)y=0

y^{^{''}} - (\frac{x}{x - 1}) y^{^{'}} + \frac{1}{x - 1} y = 0

y = e^{x + c_{1} - 1} + c_{2} x

Solution steps

y^{''} (x) - (\frac{x}{x - 1}) y^{'} (x) + \frac{1}{x - 1} y = 0

Solve using substitution:

y = e^{x + c_{1} - 1} + c_{2} x

y = e^{x + c_{1} - 1} + c_{2} x

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What is the solution for y^{''}-(x/(x-1))y^'+1/(x-1)y=0 ?
The solution for y^{''}-(x/(x-1))y^'+1/(x-1)y=0 is y=e^{x+c_{1}-1}+c_{2}x