What is the solution for y^{''}-y^'-2y=2t^2-5 ?

The solution for y^{''}-y^'-2y=2t^2-5 is y=c_{1}e^{2t}+c_{2}e^{-t}-t^2+t+1

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y^{''}-y^'-2y=2t^2-5

y^{^{''}} - y^{^{'}} - 2 y = 2 t^{2} - 5

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

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 2 t^{2} - 5

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 2^{t - 2}, y (2) = 1, y^{^{'}} (2) = - 1

y^{^{''}} + 4 y = cos (2) (x)

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

y^{^{''}} - y = \frac{2 e ^{2 x}}{e ^{2 x} + 1}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 4 sin (x)

What is the solution for y^{''}-y^'-2y=2t^2-5 ?
The solution for y^{''}-y^'-2y=2t^2-5 is y=c_{1}e^{2t}+c_{2}e^{-t}-t^2+t+1