Asked by Maily
solve this differential equation:
dy/dt=0.06y(600-y)
y(0)=10
y(t)=?
dy/dt=0.06y(600-y)
y(0)=10
y(t)=?
Answers
Answered by
Steve
y' = 0.06y(600-y)
dy / y(600-y) = .06 dx
using partial fractions,
(1/y - 1/(y-600)) dy = 600*.06 dx
ln [y/(y-600)] = 36x+c
y/(y-600) = e^(36x+c)
some fiddling and combining constants gives
y = 600e^36x/(e^36x + c)
y(0) = 10, so
10 = 600/(1+c)
c = 59
y = 600e^36x/(e^36x+59)
better double-check, making sure it fits the original equation
dy / y(600-y) = .06 dx
using partial fractions,
(1/y - 1/(y-600)) dy = 600*.06 dx
ln [y/(y-600)] = 36x+c
y/(y-600) = e^(36x+c)
some fiddling and combining constants gives
y = 600e^36x/(e^36x + c)
y(0) = 10, so
10 = 600/(1+c)
c = 59
y = 600e^36x/(e^36x+59)
better double-check, making sure it fits the original equation