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

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

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

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

y = e^{- 2 t} (2 cos (3 t) + \frac{4}{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{4}{3} sin (3 t))

y = e^{- 2 t} (2 cos (3 t) + \frac{4}{3} sin (3 t))

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

6 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

x^{^{''}} + 2 x^{^{'}} + 9 x = 0, x (0) = 0, x^{^{'}} (0) = 1

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