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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 20 cos (2 t)

y^{^{''}} + 12 y^{^{'}} + 36 y = e^{- 6 t} ln (t)

y^{^{''}} - 4 y^{^{'}} = 2 x^{2} + 5

4 y^{^{''}} + 12 y^{^{'}} + 9 = 0

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