What is the solution for y^{''}+4y^'+4y=8x ?

The solution for y^{''}+4y^'+4y=8x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+2x-2

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y^{''}+4y^'+4y=8x

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 x

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 8 x

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y^{^{'}} (0) = - 2, y (0) = 1

2 x^{^{''}} + 5 x^{^{'}} + 2 x = 4 e^{- 2 t} - t

y^{^{''}} - 9 y = \frac{9 t}{e ^{3 t}}

\frac{d ^{2} x}{d t ^{2}} - 3 \frac{d x}{d t} + 12 x = 9 t

What is the solution for y^{''}+4y^'+4y=8x ?
The solution for y^{''}+4y^'+4y=8x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+2x-2