What is the solution for y^{''}-2y^'+y=t^{-4}e^t ?

The solution for y^{''}-2y^'+y=t^{-4}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(6t^2)

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y^{''}-2y^'+y=t^{-4}e^t

y^{^{''}} - 2 y^{^{'}} + y = t^{- 4} e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t^{- 4} e^{t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

y^{^{''}} + y^{^{'}} = 10 x^{4} + 2

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

x^{^{''}} - 4 x = 3 t

\frac{d ^{2} y}{d x ^{2}} - 6 y = e^{3 x} (x^{2} + 2)

2 y^{^{''}} + y^{^{'}} + 2 y = 1, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+y=t^{-4}e^t ?
The solution for y^{''}-2y^'+y=t^{-4}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(6t^2)