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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = x sin (x)

y^{^{'''}} + y^{^{''}} + 4 y^{^{'}} + 4 y = 100 x sin (3 x)

y^{^{''}} + 4 y = sin (2 t) + t cos (t)

y^{^{''}} + 4 y^{^{'}} + 3 y = 5 e^{2 x}

y^{^{''}} - 12 y^{^{'}} + 35 y = - 40 e^{- 3 t}

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