What is the solution for x^{''}+2x^'+5x=0,x(0)=1,x^'(0)=-1 ?

The solution for x^{''}+2x^'+5x=0,x(0)=1,x^'(0)=-1 is x=e^{-t}cos(2t)

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x^{''}+2x^'+5x=0,x(0)=1,x^'(0)=-1

x^{^{''}} + 2 x^{^{'}} + 5 x = 0, x (0) = 1, x^{^{'}} (0) = - 1

x = e^{- t} cos (2 t)

Solution steps

x^{''} (t) + 2 x^{'} (t) + 5 x = 0

Solve linear ODE:

x = e^{- t} cos (2 t)

x = e^{- t} cos (2 t)

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What is the solution for x^{''}+2x^'+5x=0,x(0)=1,x^'(0)=-1 ?
The solution for x^{''}+2x^'+5x=0,x(0)=1,x^'(0)=-1 is x=e^{-t}cos(2t)