What is the solution for y^{''}-2y^'-3y=32e^{-x} ?

The solution for y^{''}-2y^'-3y=32e^{-x} is y=c_{1}e^{3x}+c_{2}e^{-x}-8xe^{-x}

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y^{''}-2y^'-3y=32e^{-x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 32 e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{- x} - 8 x e^{- x}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 32 e^{- x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - 8 x e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{- x} - 8 x e^{- x}

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

x^{^{'''}} - 4 x^{^{''}} - 3 x^{^{'}} + 18 = 0

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

y^{^{''}} + (- 16) y + 64 = - \frac{2}{3 x} e^{8 x}

y^{^{''}} + y = 2, y (0) = 4, y^{^{'}} (0) = 2

What is the solution for y^{''}-2y^'-3y=32e^{-x} ?
The solution for y^{''}-2y^'-3y=32e^{-x} is y=c_{1}e^{3x}+c_{2}e^{-x}-8xe^{-x}