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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{- 3 t}

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

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 3 e^{- x}

y^{^{''}} + 4 y^{^{'}} + 20 y = e^{- 2 t} sin (4 t)

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