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

The solution for y^{''}-8y^'+16y=t^{-8}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(42t^6)

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

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

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{42 t ^{6}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{42 t ^{6}}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{42 t ^{6}}

\int (cos^{4} (x) sin (x)) d x

\int (x^{2} + 1) (x^{3} + 3 x)^{4} d x

\int \frac{2 x + 5}{x ^{2} + 5 x} d x

f (x) = 7 + 5 x

y^{^{'}} = k \cdot y

What is the solution for y^{''}-8y^'+16y=t^{-8}e^{4t} ?
The solution for y^{''}-8y^'+16y=t^{-8}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(42t^6)