What is the solution for y^{''}+2y^'-3y=16e^{-3t} ?

The solution for y^{''}+2y^'-3y=16e^{-3t} is y=c_{1}e^t+c_{2}e^{-3t}-4te^{-3t}

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y^{''}+2y^'-3y=16e^{-3t}

y^{^{''}} + 2 y^{^{'}} - 3 y = 16 e^{- 3 t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = 16 e^{- 3 t}

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} + 196 y = 14 sin (14 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 6 sin (2 x) + 7 cos (2 x)

(D^{2} - 3 D + 2) y = 2 x^{3} - 9 x^{2} + 6 x

y^{^{''}} + 6 y^{^{'}} + 13 y = 7 cos (3 t + \frac{π}{4})

What is the solution for y^{''}+2y^'-3y=16e^{-3t} ?
The solution for y^{''}+2y^'-3y=16e^{-3t} is y=c_{1}e^t+c_{2}e^{-3t}-4te^{-3t}