What is the solution for y^{''}+4y^'+3y=6+e^{2t} ?

The solution for y^{''}+4y^'+3y=6+e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+2+1/15 e^{2t}

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y^{''}+4y^'+3y=6+e^{2t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 + e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 2 + \frac{1}{15} e^{2 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 6 + e^{2 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 2 + \frac{1}{15} e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 2 + \frac{1}{15} e^{2 t}

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 H (t)

y^{^{''}} - y^{^{'}} + 441 y = 21 sin (21 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{x} sin (2 x) + 5 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- t} \cdot cos (2 t)

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

What is the solution for y^{''}+4y^'+3y=6+e^{2t} ?
The solution for y^{''}+4y^'+3y=6+e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+2+1/15 e^{2t}