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

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

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

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

y = - 2 cos (\frac{10}{3} t)

Solution steps

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

Solve linear ODE:

y = - 2 cos (\frac{10}{3} t)

y = - 2 cos (\frac{10}{3} t)

2 y^{^{''}} + 9 y = 0

5 f^{^{''}} (t) + 50 f^{^{'}} (t) + f (t) = 0

\frac{d ^{2} y}{d x ^{2}} = - 6 y

(2000) y^{^{''}} + 18000 y^{^{'}} + 5000 y = 0

(d^{3} - 1) y = 0

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