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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 3

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 25 y = 2 x cos (x)

y^{^{''}} + 4 y = sec^{3} (2 t)

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = e^{x}

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