What is the solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=-1 ?

The solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=-1 is y=e^{-t}(3cos(3t)+2/3 sin(3t))

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y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=-1

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

x \to 2 lim (4 x^{2} - 8 x)

\frac{d}{d x} (\frac{e ^{6 x}}{x})

x \to 3 lim (x + 3)

- \frac{18}{x ^{3}}

\frac{d y}{d x} = \frac{x y + y ^{2} + x ^{2}}{x ^{2}}

What is the solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=-1 ?
The solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=-1 is y=e^{-t}(3cos(3t)+2/3 sin(3t))