What is the solution for y^{''}+4y^'+4y=3x+4 ?

The solution for y^{''}+4y^'+4y=3x+4 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+1/4

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y^{''}+4y^'+4y=3x+4

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 4

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + \frac{1}{4}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + \frac{1}{4}

(D^{2} + D - 12) y = x

y^{^{''}} - 8 y^{^{'}} + 16 y = (x - 1) e^{4 x}

y^{^{''}} + y = t - (t - 4) H (t - 2)

y^{^{''}} - 5 y^{^{'}} - 14 y = e^{x} (x + 1)

y^{^{''}} + 4 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+4y=3x+4 ?
The solution for y^{''}+4y^'+4y=3x+4 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+1/4