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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 21 y = 800 e^{- t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 13 t e^{4 t}

y^{^{''}} - 2 y^{^{'}} - 8 y = (x + 1) e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 56 cos (3 x)

(d^{2} + 6 d + 10) y = 3 e^{- 3 x} x - 2 e^{3 x} cos (x)

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