What is the solution for y^{''}-8y^'+16y=t^{-7}e^{4t} ?

The solution for y^{''}-8y^'+16y=t^{-7}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(30t^5)

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y^{''}-8y^'+16y=t^{-7}e^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 7} e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{30 t ^{5}}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = t^{- 7} e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{30 t ^{5}}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{30 t ^{5}}

x \to 1 + lim (x^{2})

x \to 5 lim (\frac{x}{x - 5})

\int \frac{1}{y ^{\frac{1}{3}}} d y

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

\frac{d}{d s} (ln (\frac{s - 1}{s ^{2} + 2 s + 5}))

What is the solution for y^{''}-8y^'+16y=t^{-7}e^{4t} ?
The solution for y^{''}-8y^'+16y=t^{-7}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(30t^5)