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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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