What is the solution for y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0 ?

The solution for y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0 is y=e^{-20t}(5cos(40t)+5/2 sin(40t))

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y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0

y^{^{''}} + 40 y^{^{'}} + 2000 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 40 y^{'} (t) + 2000 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))

y^{^{'''}} - 2 y^{^{''}} - 25 y^{^{'}} + 50 y = 0

(d^{3} + 3 d^{2} + 3 d + 1) y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} - 6 y^{^{''}} + 5 y^{^{'}} = 0

What is the solution for y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0 ?
The solution for y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0 is y=e^{-20t}(5cos(40t)+5/2 sin(40t))