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}

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