What is the solution for y^{''}-8y^'+16y=2te^{4t} ?

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

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y^{''}-8y^'+16y=2te^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 t e^{4 t}

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

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 2 t e^{4 t}

Solve linear ODE:

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

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

e^{- \frac{x}{2}}, 3

k = 1 \sum \infty \frac{x ^{k}}{k !}

\frac{d}{d x} (\frac{- 3 x ^{6} - x ^{5} + 3 x ^{4}}{x ^{2}})

x y^{^{'}} + y = y^{2}

\frac{d}{d x} (2 + x)

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