What is the solution for y^{''}+4y^'+3y=9e^{-3x} ?

The solution for y^{''}+4y^'+3y=9e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-3x}-(9e^{-3x}x)/2

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y^{''}+4y^'+3y=9e^{-3x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 9 e^{- 3 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - \frac{9 e ^{- 3 x} x}{2}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 9 e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - \frac{9 e ^{- 3 x} x}{2}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} - \frac{9 e ^{- 3 x} x}{2}

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

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{t} sin (t)

What is the solution for y^{''}+4y^'+3y=9e^{-3x} ?
The solution for y^{''}+4y^'+3y=9e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-3x}-(9e^{-3x}x)/2