What is the solution for y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7 ?

The solution for y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7 is y=e^{-2t}(4cos(3t)+1/3 sin(3t))

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y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

y = e^{- 2 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

y = e^{- 2 t} (4 cos (3 t) + \frac{1}{3} sin (3 t))

y^{^{''}} - 2 y^{^{'}} + y = 0, y (2) = 1, y^{^{'}} (2) = 2

y^{^{''}} + 14 y^{^{'}} + 51 y = 0

y^{^{'''}} - 4 y^{^{''}} - 5 y^{^{'}} = 0, y (1) = 0, y^{^{'}} (1) = 8

y^{^{''}} - 1 y^{^{'}} + 0.1 y = 0

(d^{4} - 1) y = 0

What is the solution for y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7 ?
The solution for y^{''}+4y^'+13y=0,y(0)=4,y^'(0)=-7 is y=e^{-2t}(4cos(3t)+1/3 sin(3t))