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

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

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

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

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

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

\int x x^{2} + a^{2} d x

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

\frac{\partial}{\partial x} (ln (x + 5 y + 6 z))

\frac{d}{d x} (x sec (x))

\int (cos (θ)) d θ

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