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

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

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y^{''}+2y^'=3e^{-2t}

y^{^{''}} + 2 y^{^{'}} = 3 e^{- 2 t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) = 3 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y = e^{x}

y^{^{''}} - 4 y = - 12 e^{- 2 x} + 15 cos (x) + 8 x

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 + x ^{2}}

y^{^{''}} - 4 y = (x^{2} - 9) sin (2 x)

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{- 3 x} - e^{- x}

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