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)

x \to 2 lim (3 x^{2} - 2 x + 4)

\int (x^{3} + x)^{8} (3 x^{2} + 1) d x

\int \frac{6}{2 x + 3 x ^{2}} d x

\frac{d}{d x} (\frac{e ^{x}}{9 e ^{x} + x ^{2}})

a re a - x^{2} + 10 x - 16, 2 x - 4, 2, 8

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)