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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 2 x^{^{'}} + 7 x = 6 cos (2 t)

y^{^{''}} + 36 y = - 36 sin (6 t)

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

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

y^{^{''}} + 9 y = 2 x^{2} - 5

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