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

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

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y^{''}-2y^'+y=(e^x)/((1-x)^2)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 - x ) ^{2}}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = \frac{e ^{x}}{( 1 - x ) ^{2}}

Solve linear ODE:

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

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

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