What is the solution for y^{''}-2y^'+y=t^{-8}e^t ?

The solution for y^{''}-2y^'+y=t^{-8}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(42t^6)

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y^{''}-2y^'+y=t^{-8}e^t

y^{^{''}} - 2 y^{^{'}} + y = t^{- 8} e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{42 t ^{6}}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t^{- 8} e^{t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{42 t ^{6}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{42 t ^{6}}

y^{^{''}} - 8 y^{^{'}} + 20 y = t e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{'}} - 4 y = 25 sin (x)

y^{^{''}} + 5 y^{^{'}} + 4 y = 1, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} + 400 y = 20 sin (20 t)

4 y^{^{''}} - y = x e^{\frac{x}{2}}, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+y=t^{-8}e^t ?
The solution for y^{''}-2y^'+y=t^{-8}e^t is y=c_{1}e^t+c_{2}te^t+(e^t)/(42t^6)