What is the solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-4 ?

The solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-4 is y=e^{-3t}(2cos(5t)+2/5 sin(5t))

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y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-4

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = - 4

y = e^{- 3 t} (2 cos (5 t) + \frac{2}{5} sin (5 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{- 3 t} (2 cos (5 t) + \frac{2}{5} sin (5 t))

y = e^{- 3 t} (2 cos (5 t) + \frac{2}{5} sin (5 t))

What is the solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-4 ?
The solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-4 is y=e^{-3t}(2cos(5t)+2/5 sin(5t))