What is the solution for 0.05y^{''}+2y^'+1/(0.01)y=0 ?

The solution for 0.05y^{''}+2y^'+1/(0.01)y=0 is y=e^{-20t}(c_{1}cos(40t)+c_{2}sin(40t))

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0.05y^{''}+2y^'+1/(0.01)y=0

0.05 y^{^{''}} + 2 y^{^{'}} + \frac{1}{0.01} y = 0

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

Solution steps

0.05 y^{''} (t) + 2 y^{'} (t) + \frac{1}{0.01} y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

y = e^{- 20 t} (c_{1} cos (40 t) + c_{2} sin (40 t))

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

4 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + y = 0

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

215 y^{^{''}} + 25580 y = 0

y^{^{''}} - 6 y - 16 y = 0, y (0) = 3, y^{^{'}} (0) = 4

What is the solution for 0.05y^{''}+2y^'+1/(0.01)y=0 ?
The solution for 0.05y^{''}+2y^'+1/(0.01)y=0 is y=e^{-20t}(c_{1}cos(40t)+c_{2}sin(40t))