What is the solution for y^{''}-4y^'-32y=-80t+64t^2 ?

The solution for y^{''}-4y^'-32y=-80t+64t^2 is y=c_{1}e^{8t}+c_{2}e^{-4t}-2t^2+3t-1/2

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y^{''}-4y^'-32y=-80t+64t^2

y^{^{''}} - 4 y^{^{'}} - 32 y = - 80 t + 64 t^{2}

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{1}{2}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 32 y = - 80 t + 64 t^{2}

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{1}{2}

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{1}{2}

D^{3} y = e^{3 x}

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

y^{^{''}} + 4 y^{^{'}} - 2 y = \frac{1}{x ^{2} e ^{2 x}}

- y^{^{''}} + y = e^{x}

y^{^{''}} - y^{^{'}} + y = 4, y^{^{'}} (0) = - 1, y (0) = 4

What is the solution for y^{''}-4y^'-32y=-80t+64t^2 ?
The solution for y^{''}-4y^'-32y=-80t+64t^2 is y=c_{1}e^{8t}+c_{2}e^{-4t}-2t^2+3t-1/2