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

The solution for y^{''}+2y^'=1+t^2+e^{-2t} is y=c_{1}+c_{2}e^{-2t}+(t^3)/6+(-t^2+3t)/4-(e^{-2t}t)/2

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y^{''}+2y^'=1+t^2+e^{-2t}

y^{^{''}} + 2 y^{^{'}} = 1 + t^{2} + e^{- 2 t}

y = c_{1} + c_{2} e^{- 2 t} + \frac{t ^{3}}{6} + \frac{- t ^{2} + 3 t}{4} - \frac{e ^{- 2 t} t}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) = 1 + t^{2} + e^{- 2 t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + \frac{t ^{3}}{6} + \frac{- t ^{2} + 3 t}{4} - \frac{e ^{- 2 t} t}{2}

y = c_{1} + c_{2} e^{- 2 t} + \frac{t ^{3}}{6} + \frac{- t ^{2} + 3 t}{4} - \frac{e ^{- 2 t} t}{2}

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

\frac{d}{d x} (\frac{- x}{y})

\frac{2}{x + 5}

(- 1, - 4), (5, - 4)

\frac{d}{d x} (4 x - 5 x^{\frac{7}{8}})

What is the solution for y^{''}+2y^'=1+t^2+e^{-2t} ?
The solution for y^{''}+2y^'=1+t^2+e^{-2t} is y=c_{1}+c_{2}e^{-2t}+(t^3)/6+(-t^2+3t)/4-(e^{-2t}t)/2