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

The solution for y^{''}-2y^'+y=e^xln(x) is y=c_{1}e^x+c_{2}xe^x+(2e^xx^2ln(x)-3e^xx^2)/4

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{2 e ^{x} x ^{2} ln ( x ) - 3 e ^{x} x ^{2}}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{2 e ^{x} x ^{2} ln ( x ) - 3 e ^{x} x ^{2}}{4}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{2 e ^{x} x ^{2} ln ( x ) - 3 e ^{x} x ^{2}}{4}

\frac{d}{d x} (arcsin (\frac{x}{4}))

\frac{d}{d x} (tan^{4} (3 x))

\int y^{3} e^{- y} d y

\int - \frac{14}{15} \cdot (x - 30) + 14 d x

\frac{d y}{d x} = e^{2 x - 1} y^{2}

What is the solution for y^{''}-2y^'+y=e^xln(x) ?
The solution for y^{''}-2y^'+y=e^xln(x) is y=c_{1}e^x+c_{2}xe^x+(2e^xx^2ln(x)-3e^xx^2)/4