What is the solution for y^{''}+2y^'+65y=0,y(0)=-1,y^'(0)=33 ?

The solution for y^{''}+2y^'+65y=0,y(0)=-1,y^'(0)=33 is y=e^{-t}(-cos(8t)+4sin(8t))

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y^{''}+2y^'+65y=0,y(0)=-1,y^'(0)=33

y^{^{''}} + 2 y^{^{'}} + 65 y = 0, y (0) = - 1, y^{^{'}} (0) = 33

y = e^{- t} (- cos (8 t) + 4 sin (8 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 65 y = 0

Solve linear ODE:

y = e^{- t} (- cos (8 t) + 4 sin (8 t))

y = e^{- t} (- cos (8 t) + 4 sin (8 t))

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 1, y (0) = 0

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 8

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y^{0} = 1

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 5

What is the solution for y^{''}+2y^'+65y=0,y(0)=-1,y^'(0)=33 ?
The solution for y^{''}+2y^'+65y=0,y(0)=-1,y^'(0)=33 is y=e^{-t}(-cos(8t)+4sin(8t))