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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 3 x^{^{'}} + 33 x = 10 cos (5 t)

y^{^{''}} + 4 y^{^{'}} + 3 y = 5 e^{- 3 t}

y^{^{'''}} - 2 y^{^{''}} + 5 y^{^{'}} = e^{x} cos (2 x)

(D^{2} - 3 D + 2) (D^{2} + 2 D + 1) y = x

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 t^{2}

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