What is the solution for y^{''}+7y^'+12y=2e^{-t} ?

The solution for y^{''}+7y^'+12y=2e^{-t} is y=c_{1}e^{-3t}+c_{2}e^{-4t}+1/3 e^{-t}

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y^{''}+7y^'+12y=2e^{-t}

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{- t}

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

Solution steps

y^{''} (t) + 7 y^{'} (t) + 12 y = 2 e^{- t}

Solve linear ODE:

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

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

y^{^{''}} + 16 y = sin (8 x)

2 x^{^{''}} + 3 x^{^{'}} + 5 x = 2 t

y^{^{''}} - 6 y^{^{'}} + 8 y = 8 e^{4 x}

2 y^{^{''}} + 18 y = 3 tan (3 x)

y^{^{''}} + y = e^{- 2 t} sin (t), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+7y^'+12y=2e^{-t} ?
The solution for y^{''}+7y^'+12y=2e^{-t} is y=c_{1}e^{-3t}+c_{2}e^{-4t}+1/3 e^{-t}