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

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

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

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

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))

\frac{y ^{^{''}} + y + y}{x} = 0

y^{^{''}} + 2 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{'''}} - 13 y^{^{''}} + 46 y^{^{'}} - 48 y = 0

y^{^{''}} + a y = 0, y^{^{'}} (0) = 0

y^{^{''}} + 5 y + 4 y = 0

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