What is the solution for y^{''}+6y^'+13y=0,y(0)=3 ?

The solution for y^{''}+6y^'+13y=0,y(0)=3 is y=e^{-3t}(3cos(2t)+c_{2}sin(2t))

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y^{''}+6y^'+13y=0,y(0)=3

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3

y = e^{- 3 t} (3 cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{- 3 t} (3 cos (2 t) + c_{2} sin (2 t))

y = e^{- 3 t} (3 cos (2 t) + c_{2} sin (2 t))

2 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

y^{^{''}} + 1 y = 0, y (π^{2}) = 3, y^{^{'}} (π^{2}) = - 1

- 2 y^{^{''}} + 4 y^{^{'}} = 0

8 x^{^{''}} + 20 x^{^{'}} + 32 x = 0

(D^{4} + 18 D^{2} + 81) y = 0

What is the solution for y^{''}+6y^'+13y=0,y(0)=3 ?
The solution for y^{''}+6y^'+13y=0,y(0)=3 is y=e^{-3t}(3cos(2t)+c_{2}sin(2t))