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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

t \to t lim (3 t^{\frac{1}{3}})

\frac{d}{d x} ((2 x + 1)^{5} (x^{3} - x + 1)^{4})

\int 2 - 2 cos (x) d x

4 x^{\frac{5}{4}} + 2 x^{\frac{3}{2}} + 4 x

\frac{3}{z ^{2}}

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