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

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

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

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

y = e^{- t} (2 cos (2 t) + \frac{1}{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (2 cos (2 t) + \frac{1}{2} sin (2 t))

y = e^{- t} (2 cos (2 t) + \frac{1}{2} sin (2 t))

n \to \infty lim (\frac{1}{x ^{n - 1}})

\int_{1}^{3} x d x

\frac{d}{d x} (\frac{1}{e ^{x} + 2})

\int \frac{( x ^{2} - x + 1 )}{x ^{2} + x} d x

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