What is the solution for y^{''}+6y^'+13y=0,y(0)=-3,y^'(0)= 1/3 ?

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

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y^{''}+6y^'+13y=0,y(0)=-3,y^'(0)= 1/3

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = \frac{1}{3}

y = e^{- 3 t} (- 3 cos (2 t) - \frac{13}{3} 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{13}{3} sin (2 t))

y = e^{- 3 t} (- 3 cos (2 t) - \frac{13}{3} sin (2 t))

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