What is the solution for y^{''}-2y^'+y=(e^t)/(t^3) ?

The solution for y^{''}-2y^'+y=(e^t)/(t^3) is y=c_{1}e^t+c_{2}te^t+(e^t)/(2t)

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y^{''}-2y^'+y=(e^t)/(t^3)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t ^{3}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{2 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = \frac{e ^{t}}{t ^{3}}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{2 t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{2 t}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 3

y^{^{'''}} + 4 y^{^{''}} + 5 y^{^{'}} + 2 y = 10 cos (t)

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{2 x}

y^{^{''}} - y^{^{'}} - 20 y = 27 e^{- 4 x}

What is the solution for y^{''}-2y^'+y=(e^t)/(t^3) ?
The solution for y^{''}-2y^'+y=(e^t)/(t^3) is y=c_{1}e^t+c_{2}te^t+(e^t)/(2t)