What is the solution for y^{''}+4y^'+9y=0,y(0)=1,y^'(0)=13 ?

The solution for y^{''}+4y^'+9y=0,y(0)=1,y^'(0)=13 is y=e^{-2t}(cos(sqrt(5)t)+3sqrt(5)sin(sqrt(5)t))

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y^{''}+4y^'+9y=0,y(0)=1,y^'(0)=13

y^{^{''}} + 4 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 13

y = e^{- 2 t} (cos (5 t) + 35 sin (5 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- 2 t} (cos (5 t) + 35 sin (5 t))

y = e^{- 2 t} (cos (5 t) + 35 sin (5 t))

9 (5^{x})

y^{^{'}} = 9 x y

x \to 2 - lim (x^{2} + 3)

\frac{d y}{d x} = (\frac{2 y + 1}{4 x - 3})^{2}

\frac{d}{d x} ((x^{2} + 4)^{5})

What is the solution for y^{''}+4y^'+9y=0,y(0)=1,y^'(0)=13 ?
The solution for y^{''}+4y^'+9y=0,y(0)=1,y^'(0)=13 is y=e^{-2t}(cos(sqrt(5)t)+3sqrt(5)sin(sqrt(5)t))