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

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

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

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

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

Solution steps

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

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}

\frac{d y}{d x} + \frac{3 y}{x} = \frac{e ^{x}}{x ^{3}}

\int - 2 x + 4 d x

\frac{\partial}{\partial x} (- 8 x)

\int (x + 2)^{2} d x

\int 6 x^{3} e^{x^{4}} d x

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