What is the solution for x^{''}+2x^'+2=0,x(0)=0,x^'(0)=1 ?

The solution for x^{''}+2x^'+2=0,x(0)=0,x^'(0)=1 is x=1-e^{-2t}-t

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x^{''}+2x^'+2=0,x(0)=0,x^'(0)=1

x^{^{''}} + 2 x^{^{'}} + 2 = 0, x (0) = 0, x^{^{'}} (0) = 1

x = 1 - e^{- 2 t} - t

Solution steps

x^{''} (t) + 2 x^{'} (t) + 2 = 0

Solve linear ODE:

x = 1 - e^{- 2 t} - t

x = 1 - e^{- 2 t} - t

x \to 0 lim (\frac{1}{x ^{3} - 7 x})

a re a f (x) = 2 x^{3} + 3 x^{2} - 5 x, - 4, 3

\frac{\partial}{\partial x} (x^{3} ln (yz))

\int x x - 12 d x

x \frac{d v}{d x} = \frac{2}{v} + v

What is the solution for x^{''}+2x^'+2=0,x(0)=0,x^'(0)=1 ?
The solution for x^{''}+2x^'+2=0,x(0)=0,x^'(0)=1 is x=1-e^{-2t}-t