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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x^{2}

0.5 y^{^{''}} + y^{^{'}} + 4 y = 10 sin (2 x)

y^{^{''}} - y^{^{'}} = \frac{e ^{x}}{e ^{x} + 1}

y^{^{''}} - 4 y = x sin (x)

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

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