What is the solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-12 ?

The solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-12 is y=e^{-2t}(3cos(3t)-2sin(3t))

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y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-12

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 12

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

Solution steps

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

Solve linear ODE:

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

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

x \to 2 lim (x - 5)

\int_{0}^{2} \frac{1}{( 2 x - 1 ) ^{\frac{2}{3}}} d x

\frac{\partial}{\partial x} (x^{4} + 3 y^{2} - x)

\frac{x ^{2} - 4}{x - 2}

\frac{d}{d x} (x + y + 2)

What is the solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-12 ?
The solution for y^{''}+4y^'+13y=0,y(0)=3,y^'(0)=-12 is y=e^{-2t}(3cos(3t)-2sin(3t))