What is the solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=5 ?

The solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=5 is y=e^{-2t}(2cos(3t)+3sin(3t))

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y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=5

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 5

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

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

y^{^{''}} + \frac{0}{3} y^{^{'}} + \frac{12}{3} y = 0

3 y^{^{'''}} - 2 y^{^{''}} - 37 y^{^{'}} - 12 y = 0

- y^{^{''}} + 4 y^{^{'}} + 5 y = 0

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 1

(d^{3} + 9 d^{2} + 27 d + 27) y = 0

What is the solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=5 ?
The solution for y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=5 is y=e^{-2t}(2cos(3t)+3sin(3t))