What is the solution for y^{''}-4y^'+13y=0,y(0)=1,y^'(0)=-4 ?

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

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y^{''}-4y^'+13y=0,y(0)=1,y^'(0)=-4

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial z} (\frac{x y}{x + y + z})

\frac{d}{d x} (\frac{cos ( x )}{4 x})

x \to 1 - lim (x^{3} - 4)

\int 5 - 3 (1 + x^{2})^{- 1} d x

y = (arctan (8 x))^{2}

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