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

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

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

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

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 + 4

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^{^{''}} + 4 y^{^{'}} + 5 y = e^{- t}

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

x^{^{''}} + 4 x = 4 t

y^{^{''}} + 16 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2} + 8 x + 9

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