What is the solution for y^{''}+4y^'+3y=2e^{-4x} ?

The solution for y^{''}+4y^'+3y=2e^{-4x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+2/3 e^{-4x}

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2}{3} e^{- 4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2}{3} e^{- 4 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{2}{3} e^{- 4 x}

What is the solution for y^{''}+4y^'+3y=2e^{-4x} ?
The solution for y^{''}+4y^'+3y=2e^{-4x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+2/3 e^{-4x}