What is the solution for y^{''}+6y^'+5y=e^{-3x} ?

The solution for y^{''}+6y^'+5y=e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-5x}-1/4 e^{-3x}

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y^{''}+6y^'+5y=e^{-3x}

y^{^{''}} + 6 y^{^{'}} + 5 y = e^{- 3 x}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} - \frac{1}{4} e^{- 3 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 5 y = e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 5 x} - \frac{1}{4} e^{- 3 x}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} - \frac{1}{4} e^{- 3 x}

2 y^{^{''}} + 3 y^{^{'}} + y = x^{3} + 3 sin (x)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 54 + 18

d, d (x) = e^{2 x^{3} + 4}

x^{^{''}} + 4 x = sin (3 t)

(D^{2} - 4 D + 3) y = 9 x^{2} - 36 x + 37

What is the solution for y^{''}+6y^'+5y=e^{-3x} ?
The solution for y^{''}+6y^'+5y=e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-5x}-1/4 e^{-3x}