What is the solution for y^{''}-2y^'+y=t^{-7}e^t ?

The solution for y^{''}-2y^'+y=t^{-7}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(30t^5)

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y^{''}-2y^'+y=t^{-7}e^t

y^{^{''}} - 2 y^{^{'}} + y = t^{- 7} e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{30 t ^{5}}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t^{- 7} e^{t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{30 t ^{5}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{30 t ^{5}}

y^{^{''}} - y - 2 = 0, y (0) = 3

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

y^{^{''}} + 3 y^{^{'}} - 4 = 0

y^{^{''}} + 4 y + 4 = x^{2} + 2 sin (x) cos (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{12 e ^{x}}{x ^{3}}

What is the solution for y^{''}-2y^'+y=t^{-7}e^t ?
The solution for y^{''}-2y^'+y=t^{-7}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(30t^5)