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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (x^{2 - 2 x})

y = \frac{ln ( x )}{e ^{x}}

\int csc (6 x) d x

\frac{2 s}{s ^{2} + 2 s + 1}

\int_{0}^{\frac{π}{4}} (sin (x)) d x

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