What is the solution for 2y^{''}+12y^'+68y=0,y(0)=3,y^'(0)=0 ?

The solution for 2y^{''}+12y^'+68y=0,y(0)=3,y^'(0)=0 is y=e^{-3t}(3cos(5t)+9/5 sin(5t))

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2y^{''}+12y^'+68y=0,y(0)=3,y^'(0)=0

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y = e^{- 3 t} (3 cos (5 t) + \frac{9}{5} sin (5 t))

Solution steps

2 y^{''} (t) + 12 y^{'} (t) + 68 y = 0

Solve linear ODE:

y = e^{- 3 t} (3 cos (5 t) + \frac{9}{5} sin (5 t))

y = e^{- 3 t} (3 cos (5 t) + \frac{9}{5} sin (5 t))

x^{^{'''}} + 7 x^{^{''}} - 5 x^{^{'}} - 75 x = 0

y^{^{''}} - 14 y^{^{'}} + 5 y = 0

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

y^{^{''}} - 2 y^{^{'}} - 17 y = 0

5 y^{^{''}} - 11 y^{^{'}} + 2 y = 0, y (0) = - 5, y^{^{'}} (0) = - 6.4

What is the solution for 2y^{''}+12y^'+68y=0,y(0)=3,y^'(0)=0 ?
The solution for 2y^{''}+12y^'+68y=0,y(0)=3,y^'(0)=0 is y=e^{-3t}(3cos(5t)+9/5 sin(5t))