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

The solution for x^{''}+2x^'+2x=2 is x=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+1

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

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

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

Solution steps

x^{''} (t) + 2 x^{'} (t) + 2 x = 2

Solve linear ODE:

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

\frac{d}{d x} (- 2 ln (cos (x)))

\int x \cdot e^{2 x^{2}} d x

y = \frac{x}{x + 5}

\int_{1}^{2} - x^{2} + 3 x - 2 d x

2 x^{2} \frac{d y}{d x} + 2 x y = y^{3}

What is the solution for x^{''}+2x^'+2x=2 ?
The solution for x^{''}+2x^'+2x=2 is x=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+1