What is the solution for y^{''}+3y^'-4y=-32t^2 ?

The solution for y^{''}+3y^'-4y=-32t^2 is y=c_{1}e^t+c_{2}e^{-4t}+8t^2+12t+13

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y^{''}+3y^'-4y=-32t^2

y^{^{''}} + 3 y^{^{'}} - 4 y = - 32 t^{2}

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = - 32 t^{2}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e t

y^{^{''}} + 2 y^{^{'}} + y = 2 cos (2 x) + 3 x + 2 + e^{x}

y^{^{''}} - 2 y^{^{'}} + 17 y = 289 x + 9

y^{^{''}} + y^{^{'}} = (2 t)

y^{^{''}} + 3 y + 2 = 0, y (2) = - 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+3y^'-4y=-32t^2 ?
The solution for y^{''}+3y^'-4y=-32t^2 is y=c_{1}e^t+c_{2}e^{-4t}+8t^2+12t+13