What is the solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=-8 ?

The solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=-8 is y=e^{-3t}(3cos(2t)+1/2 sin(2t))

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y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=-8

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

y = e^{- 3 t} (3 cos (2 t) + \frac{1}{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) + \frac{1}{2} sin (2 t))

y = e^{- 3 t} (3 cos (2 t) + \frac{1}{2} sin (2 t))

16 y^{^{''}} + 56 y^{^{'}} + 130 y = 0, y (1) = 2, y^{^{'}} (1) = 5

y^{^{''}} + 4 y = 0, y (0) = 1

y^{^{''}} - 2 y + 10 y = 0

y^{^{''}} - 324 y = 0

49 y^{^{''}} + 56 y^{^{'}} + 16 y = 0

What is the solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=-8 ?
The solution for y^{''}+6y^'+13y=0,y(0)=3,y^'(0)=-8 is y=e^{-3t}(3cos(2t)+1/2 sin(2t))