What is the solution for y^{''}+4y^'+4y=6x+7 ?

The solution for y^{''}+4y^'+4y=6x+7 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/2+1/4

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y^{''}+4y^'+4y=6x+7

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 x + 7

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 6 x + 7

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} + 4 y = cos (x)

y^{^{''}} + 9 y = 2 e^{3 t} + 2 cos (2 t)

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

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

What is the solution for y^{''}+4y^'+4y=6x+7 ?
The solution for y^{''}+4y^'+4y=6x+7 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/2+1/4