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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = 12 e^{3 t}, y (0) = - 1, y^{^{'}} (0) = 0

2 y^{^{''}} - 4 y^{^{'}} + 2 y = e^{x} x^{- 1}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{2}

(D^{2} + 2 D + 1) y = e^{- 2 x} + 1

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

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