What is the solution for y^{''}+3y^'-40y=-132t+80t^2 ?

The solution for y^{''}+3y^'-40y=-132t+80t^2 is y=c_{1}e^{5t}+c_{2}e^{-8t}-2t^2+3t+1/8

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y^{''}+3y^'-40y=-132t+80t^2

y^{^{''}} + 3 y^{^{'}} - 40 y = - 132 t + 80 t^{2}

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} - 2 t^{2} + 3 t + \frac{1}{8}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 40 y = - 132 t + 80 t^{2}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} - 2 t^{2} + 3 t + \frac{1}{8}

y = c_{1} e^{5 t} + c_{2} e^{- 8 t} - 2 t^{2} + 3 t + \frac{1}{8}

y^{^{''}} + 8 y^{^{'}} + 12 = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 e^{t} + 2 t + 2

y^{^{'''}} = 3 y^{^{''}} + 9 e^{3} x + 5 cos (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 sin (x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = 10 e^{-} 1, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+3y^'-40y=-132t+80t^2 ?
The solution for y^{''}+3y^'-40y=-132t+80t^2 is y=c_{1}e^{5t}+c_{2}e^{-8t}-2t^2+3t+1/8