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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{x x ^{2} + 1}{( 2 x + 1 ) ^{2}})

\frac{d}{d x} (x^{- 6 c o s (x)})

\frac{\partial}{\partial x} (sin (x - y) + (x - y))

\int e^{\frac{x}{l}} d x

\int (x^{\frac{4}{3}} - 3 x^{\frac{5}{2}}) d x

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