What is the solution for y^{''}+4y^'+4y=3x+6 ?

The solution for y^{''}+4y^'+4y=3x+6 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+3/4

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y^{''}+4y^'+4y=3x+6

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 6

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 3 x + 6

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 8 y = 3 e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 5 y = (4 x^{2} - 7 x + 0) e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6

y^{^{''''}} - 2 y^{^{''}} + y = 5 sin (2 x)

y^{^{''}} + 3 y = e^{t}

What is the solution for y^{''}+4y^'+4y=3x+6 ?
The solution for y^{''}+4y^'+4y=3x+6 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+3/4