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

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

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

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

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

Solution steps

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

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}

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