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

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

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

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

y = e^{- t} (3 cos (3 t) + \frac{1}{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{1}{3} sin (3 t))

y = e^{- t} (3 cos (3 t) + \frac{1}{3} sin (3 t))

y^{^{''}} + 3 \cdot y^{^{'}} + 2 \cdot y = 0, y (0) = 1, y^{^{'}} (0) = 1

3 y^{^{''}} + 24 y^{^{'}} + 48 y = 0, y (0) = 8, y^{^{'}} (0) = - 33

y^{^{''}} - 4 y^{^{'}} + 29 y = 0, y (0) = 6, y^{^{'}} (0) = - 3

6 y^{^{''}} - 13 y^{^{'}} + 6 y = 0

y^{^{''}} + 5 y^{^{'}} + 8 y = 0

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