What is the solution for y^{''}+5y^'+6y=3e^{-2t} ?

The solution for y^{''}+5y^'+6y=3e^{-2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+3te^{-2t}

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y^{''}+5y^'+6y=3e^{-2t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 e^{- 2 t}

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

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 3 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 3 + 4 e^{- t}

y^{^{''}} + 9 y = 18 t, y (\frac{π}{2}) = 0, y (0) = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{- 2} e^{- 2 x}

y^{^{''}} + 16 y = sin (5 x)

4 y^{^{''}} - 4 y^{^{'}} - 35 y = cos (2 x)

What is the solution for y^{''}+5y^'+6y=3e^{-2t} ?
The solution for y^{''}+5y^'+6y=3e^{-2t} is y=c_{1}e^{-2t}+c_{2}e^{-3t}+3te^{-2t}