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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 25 y = x^{2} e^{x} - x e^{5 x}

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

y^{^{''}} + 16 y = 4 cos (4 x) + 6 sin (4 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = e - t, y (4) = 1, y^{^{'}} (4) = 3

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