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

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

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y^{''}-y^'-2y=4t^2

y^{^{''}} - y^{^{'}} - 2 y = 4 t^{2}

y = c_{1} e^{2 t} + c_{2} e^{- t} - 2 t^{2} + 2 t - 3

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - 2 t^{2} + 2 t - 3

y = c_{1} e^{2 t} + c_{2} e^{- t} - 2 t^{2} + 2 t - 3

\frac{d ^{2} y}{d x ^{2}} + 25 y = 5 x^{2} + x

25 y^{^{''}} - 9 y = sin (x)

y^{^{''}} + 2 y^{^{'}} - 24 y = 6 x

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t)

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

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