What is the solution for y^{''}+4y^'+4y=2x ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 34 y = 30 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 9 y = x e^{x}

y^{^{''''}} - 16 y = x e^{2 x}

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

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

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