What is the solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=13 ?

The solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=13 is y=e^{-2t}(2cos(3t)+17/3 sin(3t))

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y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=13

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 13

y = e^{- 2 t} (2 cos (3 t) + \frac{17}{3} sin (3 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (3 t) + \frac{17}{3} sin (3 t))

y = e^{- 2 t} (2 cos (3 t) + \frac{17}{3} sin (3 t))

16 y^{^{'''}} + 24 y^{^{''}} + 9 y^{^{'}} = 0

121 y^{^{''}} - 4 y = 0

64 y^{^{''}} + 176 y^{^{'}} + 121 y = 0

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

y^{^{'''}} - 10 y^{^{''}} + 25 y^{^{'}} = 0

What is the solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=13 ?
The solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=13 is y=e^{-2t}(2cos(3t)+17/3 sin(3t))