What is the solution for y^{''}+4y^'+4y=2x+6 ?

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

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

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

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 2 x + 6

Solve linear ODE:

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

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

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

\int 0.06 \cdot e^{\frac{t}{300}} d t

\frac{d}{d x} (ln (\frac{2 x + 1}{x - 1}))

n = 1 \sum \infty \frac{7}{( n + 1 ) !}

\frac{d}{d x} (x^{2} 6 x - 7)

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