What is the solution for y^{''}+2y^'+10y=0,y(0)=-3,y^'(0)=-4 ?

The solution for y^{''}+2y^'+10y=0,y(0)=-3,y^'(0)=-4 is y=e^{-t}(-3cos(3t)-7/3 sin(3t))

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y^{''}+2y^'+10y=0,y(0)=-3,y^'(0)=-4

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = - 3, y^{^{'}} (0) = - 4

y = e^{- t} (- 3 cos (3 t) - \frac{7}{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{7}{3} sin (3 t))

y = e^{- t} (- 3 cos (3 t) - \frac{7}{3} sin (3 t))

y^{^{''}} + 3 y = 0, y^{0} = 1, y^{^{'}} 0 = 3

\frac{1}{2} y^{^{''}} + 6 y^{^{'}} + 26 y = 0

y^{^{''}} - 9 y^{^{'}} = 0, y (0) = e^{3 x}

y^{^{''}} - 2 y^{^{'}} + 50 y = 0

y^{^{'''}} + 4 y^{^{''}} - 15 y^{^{'}} - 18 y = 0

What is the solution for y^{''}+2y^'+10y=0,y(0)=-3,y^'(0)=-4 ?
The solution for y^{''}+2y^'+10y=0,y(0)=-3,y^'(0)=-4 is y=e^{-t}(-3cos(3t)-7/3 sin(3t))