What is the solution for y^{''}+y^'-12y=-40t+24t^2 ?

The solution for y^{''}+y^'-12y=-40t+24t^2 is y=c_{1}e^{3t}+c_{2}e^{-4t}-2t^2+3t-1/12

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y^{''}+y^'-12y=-40t+24t^2

y^{^{''}} + y^{^{'}} - 12 y = - 40 t + 24 t^{2}

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

Solution steps

y^{''} (t) + y^{'} (t) - 12 y = - 40 t + 24 t^{2}

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + 4 y = e^{t} + e^{- t}

(d^{2} + 2 d + 2) y = 3 e^{x} sin (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = x - e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 4 - e^{t}, y (0) = 1, y (1) = 0

y^{^{''}} + y = 2 x^{2}

What is the solution for y^{''}+y^'-12y=-40t+24t^2 ?
The solution for y^{''}+y^'-12y=-40t+24t^2 is y=c_{1}e^{3t}+c_{2}e^{-4t}-2t^2+3t-1/12