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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y^{^{'''}} + 4 y^{^{''}} - 31 y^{^{'}} - 70 y = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 8

4 y^{^{''}} + 8 y^{^{'}} = 0

144 y^{^{'''}} + 264 y^{^{''}} + 121 y^{^{'}} = 0

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