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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 0

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

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

\int_{1}^{e^{3}} \frac{ln ^{3} ( x ^{2} )}{x} d x

\frac{x}{1 - x ^{2}}

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