What is the solution for y^{''}-y^'-20y=-56t+40t^2 ?

The solution for y^{''}-y^'-20y=-56t+40t^2 is y=c_{1}e^{5t}+c_{2}e^{-4t}-2t^2+3t-7/20

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y^{''}-y^'-20y=-56t+40t^2

y^{^{''}} - y^{^{'}} - 20 y = - 56 t + 40 t^{2}

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{7}{20}

Solution steps

y^{''} (t) - y^{'} (t) - 20 y = - 56 t + 40 t^{2}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{7}{20}

y = c_{1} e^{5 t} + c_{2} e^{- 4 t} - 2 t^{2} + 3 t - \frac{7}{20}

What is the solution for y^{''}-y^'-20y=-56t+40t^2 ?
The solution for y^{''}-y^'-20y=-56t+40t^2 is y=c_{1}e^{5t}+c_{2}e^{-4t}-2t^2+3t-7/20