What is the solution for y^{''}-2y^'-24y=-64t+48t^2 ?

The solution for y^{''}-2y^'-24y=-64t+48t^2 is y=c_{1}e^{6t}+c_{2}e^{-4t}-2t^2+3t-5/12

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y^{''}-2y^'-24y=-64t+48t^2

y^{^{''}} - 2 y^{^{'}} - 24 y = - 64 t + 48 t^{2}

y = c_{1} e^{6 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{5}{12}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 24 y = - 64 t + 48 t^{2}

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{5}{12}

y = c_{1} e^{6 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{5}{12}

y^{^{''}} - 2 y^{^{'}} + y = 13 e^{x}

y^{^{''}} - y^{^{'}} = e^{x} + 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 t

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 t e^{2 t}, y^{^{'}} (0) = - 3

y^{^{''''}} + y = cos (x)

What is the solution for y^{''}-2y^'-24y=-64t+48t^2 ?
The solution for y^{''}-2y^'-24y=-64t+48t^2 is y=c_{1}e^{6t}+c_{2}e^{-4t}-2t^2+3t-5/12