What is the solution for y^{''}+4y^'+3y=30e^{2t} ?

The solution for y^{''}+4y^'+3y=30e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+2e^{2t}

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y^{''}+4y^'+3y=30e^{2t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 30 e^{2 t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 30 e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} + 25 y = 14 cos (5 x) + 21 sin (5 x)

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 4 sin (x))

y^{^{''}} - y^{^{'}} = 5 e^{2 x}

y^{^{''}} - y - 2 y = 1 + t

What is the solution for y^{''}+4y^'+3y=30e^{2t} ?
The solution for y^{''}+4y^'+3y=30e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+2e^{2t}