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

x^{^{''}} + 2 x^{^{'}} + 5 x = 12 cos (2 t) + 3 sin (2 t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 2 cos (x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 4

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 5

- 4 y^{^{''}} + 9 y = 15

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