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

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

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

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

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

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

\int (2 x^{2} + 1)^{2} d x

\frac{d}{d x} (x^{2} - 1)

\int 4 x \cdot ln (x) d x

f (x) = (\frac{x + 1}{x - 3})^{2}, at x = 7

\frac{\partial}{\partial x} (x^{2} y^{2} cos (x))

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