What is the solution for y^{''}+4y^'+3y=8e^{-2x} ?

The solution for y^{''}+4y^'+3y=8e^{-2x} is y=c_{1}e^{-x}+c_{2}e^{-3x}-8e^{-2x}

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - 8 e^{- 2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - 8 e^{- 2 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - 8 e^{- 2 x}

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

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{- x}

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 12 e^{2 x} + 24 x^{2}

\frac{d ^{2} y}{d t ^{2}} - y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

a y^{^{''}} - 2 a y^{^{'}} - 3 a y = x^{2}

What is the solution for y^{''}+4y^'+3y=8e^{-2x} ?
The solution for y^{''}+4y^'+3y=8e^{-2x} is y=c_{1}e^{-x}+c_{2}e^{-3x}-8e^{-2x}