What is the solution for 0.05y^{''}+2y^'+100y=0,y(0)=5,y^'(0)=0 ?

The solution for 0.05y^{''}+2y^'+100y=0,y(0)=5,y^'(0)=0 is y=e^{-20t}(5cos(40t)+5/2 sin(40t))

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0.05y^{''}+2y^'+100y=0,y(0)=5,y^'(0)=0

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

Solution steps

0.05 y^{''} (t) + 2 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

(D^{4} - 81) y = 0

y^{^{''}} + y = 0, y (0) = 3, y (1) = - 3

y^{^{''}} - 12 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = 16

y^{^{''}} - π^{2} y = 0

y^{^{'''}} - 19 y^{^{''}} + 110 y^{^{'}} - 200 y = 0

What is the solution for 0.05y^{''}+2y^'+100y=0,y(0)=5,y^'(0)=0 ?
The solution for 0.05y^{''}+2y^'+100y=0,y(0)=5,y^'(0)=0 is y=e^{-20t}(5cos(40t)+5/2 sin(40t))