What is the solution for y^{''}-8y^'+16y=t^{-5}e^{4t} ?

The solution for y^{''}-8y^'+16y=t^{-5}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(12t^3)

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y^{''}-8y^'+16y=t^{-5}e^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 5} e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{12 t ^{3}}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = t^{- 5} e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{12 t ^{3}}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{12 t ^{3}}

x^{^{''}} + 7 x^{^{'}} + 12 x = e^{2 t} + cos (4 t)

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

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 8 - e^{x} + e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = - 4 e^{x} + 3 x^{2}

y^{^{''}} + 3 y^{^{'}} + 4 y = 8 x + 2

What is the solution for y^{''}-8y^'+16y=t^{-5}e^{4t} ?
The solution for y^{''}-8y^'+16y=t^{-5}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(12t^3)