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

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

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

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

x = e^{- t} sin (t)

Solution steps

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

Solve linear ODE:

x = e^{- t} sin (t)

x = e^{- t} sin (t)

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