What is the solution for y^{''}+2y^'+2y=1+x ?

The solution for y^{''}+2y^'+2y=1+x is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+x/2

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y^{''}+2y^'+2y=1+x

y^{^{''}} + 2 y^{^{'}} + 2 y = 1 + x

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) + 2 y = 1 + x

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x}{2}

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x}{2}

y^{^{''}} - 8 y^{^{'}} + 16 y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} = sin (4 x)

y^{^{''}} + 4 y = 4 cos (x) + 3 sin (x) - 8

16 y^{^{''}} - 40 y + 25 = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 12

What is the solution for y^{''}+2y^'+2y=1+x ?
The solution for y^{''}+2y^'+2y=1+x is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+x/2