What is the solution for y^{''}+y^'=-8e^{-x}-4 ?

The solution for y^{''}+y^'=-8e^{-x}-4 is y=c_{1}+c_{2}e^{-x}+8xe^{-x}-4x

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y^{''}+y^'=-8e^{-x}-4

y^{^{''}} + y^{^{'}} = - 8 e^{- x} - 4

y = c_{1} + c_{2} e^{- x} + 8 x e^{- x} - 4 x

Solution steps

y^{''} (x) + y^{'} (x) = - 8 e^{- x} - 4

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + 8 x e^{- x} - 4 x

y = c_{1} + c_{2} e^{- x} + 8 x e^{- x} - 4 x

What is the solution for y^{''}+y^'=-8e^{-x}-4 ?
The solution for y^{''}+y^'=-8e^{-x}-4 is y=c_{1}+c_{2}e^{-x}+8xe^{-x}-4x