What is the solution for y^{''}+3y^'+2y=10e^{3t} ?

The solution for y^{''}+3y^'+2y=10e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/2 e^{3t}

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y^{''}+3y^'+2y=10e^{3t}

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 e^{3 t}

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 10 e^{3 t}

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 3 y = x e^{x}

3 y^{^{''}} + 8 y^{^{'}} + 4 y = 8 x^{2} - 8 x + 2 + e^{- 2 x}

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

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

y^{^{''}} + 2 y^{^{'}} + y = t, y (0) = 1, y^{^{'}} (0) = 2

What is the solution for y^{''}+3y^'+2y=10e^{3t} ?
The solution for y^{''}+3y^'+2y=10e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/2 e^{3t}