What is the solution for y^{''}+3y^'-28y=-96t+56t^2 ?

The solution for y^{''}+3y^'-28y=-96t+56t^2 is y=c_{1}e^{4t}+c_{2}e^{-7t}-2t^2+3t+5/28

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y^{''}+3y^'-28y=-96t+56t^2

y^{^{''}} + 3 y^{^{'}} - 28 y = - 96 t + 56 t^{2}

y = c_{1} e^{4 t} + c_{2} e^{- 7 t} - 2 t^{2} + 3 t + \frac{5}{28}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 28 y = - 96 t + 56 t^{2}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 7 t} - 2 t^{2} + 3 t + \frac{5}{28}

y = c_{1} e^{4 t} + c_{2} e^{- 7 t} - 2 t^{2} + 3 t + \frac{5}{28}

θ^{^{''}} (t) - θ (t) = t

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

y^{^{''}} + 7 y^{^{'}} - 8 y = 13 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} - 11 y^{^{''}} + 35 y^{^{'}} - 25 = 0

\frac{d ^{2} y}{d t ^{2}} + 2 y = cos (7 t)

What is the solution for y^{''}+3y^'-28y=-96t+56t^2 ?
The solution for y^{''}+3y^'-28y=-96t+56t^2 is y=c_{1}e^{4t}+c_{2}e^{-7t}-2t^2+3t+5/28