What is the solution for y^{''}-8y^'+16y=16t^6e^{4t} ?

The solution for y^{''}-8y^'+16y=16t^6e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(2e^{4t}t^8)/7

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y^{''}-8y^'+16y=16t^6e^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 t^{6} e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{2 e ^{4 t} t ^{8}}{7}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 16 t^{6} e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{2 e ^{4 t} t ^{8}}{7}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{2 e ^{4 t} t ^{8}}{7}

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 x + e^{- 2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{5 x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 5 y = 20 sin (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x + 4

y^{^{''}} + 4 y = 6 sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

What is the solution for y^{''}-8y^'+16y=16t^6e^{4t} ?
The solution for y^{''}-8y^'+16y=16t^6e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(2e^{4t}t^8)/7