What is the solution for y^{''}-2y^'+y=(e^x)/(x^5) ?

The solution for y^{''}-2y^'+y=(e^x)/(x^5) is y=c_{1}e^x+c_{2}xe^x+(e^x)/(12x^3)

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y^{''}-2y^'+y=(e^x)/(x^5)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{5}}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x}}{12 x ^{3}}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = \frac{e ^{x}}{x ^{5}}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x}}{12 x ^{3}}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x}}{12 x ^{3}}

y^{^{''}} - 4 y = 3 x e^{2 x}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}

y^{^{'''}} - y^{^{''}} = x

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 12

What is the solution for y^{''}-2y^'+y=(e^x)/(x^5) ?
The solution for y^{''}-2y^'+y=(e^x)/(x^5) is y=c_{1}e^x+c_{2}xe^x+(e^x)/(12x^3)