What is the solution for y^{''}+6y^'+13y=0,y(0)=2,y^'(0)=-5 ?

The solution for y^{''}+6y^'+13y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(2t)+1/2 sin(2t))

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y^{''}+6y^'+13y=0,y(0)=2,y^'(0)=-5

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

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

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

x^{^{''}} + a x^{^{'}} = 0

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} - 20 y = 0

y^{^{''}} + 40 y^{^{'}} + 2000 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{'''}} - 2 y^{^{''}} - 25 y^{^{'}} + 50 y = 0

(d^{3} + 3 d^{2} + 3 d + 1) y = 0

What is the solution for y^{''}+6y^'+13y=0,y(0)=2,y^'(0)=-5 ?
The solution for y^{''}+6y^'+13y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(2t)+1/2 sin(2t))