anonymous
  • anonymous
Suppose that a population grows according to the unlimited growth model described by the differential equation \frac(dy)(dt) = 0.15 y and we are given the initial condition y(0) = 600. Find the size of the population at time t = 5. (Okay to round your answer to closest whole number.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ dy }{ y }=0.15 dt\] integrate both sides
anonymous
  • anonymous
could you explain further please?
zepdrix
  • zepdrix
What are you confused about maria? :) Are you familiar with `Separation of Variables`?

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anonymous
  • anonymous
@mariamka$$\frac{dy}{dt}=0.15y\\\frac1y\frac{dy}{dt}=0.15$$integrate both sides by \(t\):$$\int\frac1y\frac{dy}{dt}\,dt=\int 0.15\,dt$$recognize the left-hand integrand as the chain rule i.e. $$\frac1y\frac{dy}{dt}=\frac{d}{dt}[\log y]$$so:$$\int\frac{d}{dt}[\log y]\,dt=0.15\int \,dt\\\log y=0.15t+c_1\\y=e^{0.15t+c_1}=e^{c_1}e^{0.15t}=c_2e^{0.15t}$$ where \(c_2=e^{c_1}\)
anonymous
  • anonymous
now observe our initial condition \(y(0)=600\) must be satisfied so:$$y(t)=c_2e^{0.15t}\\y(0)=c_2e^{0.15(0)}=c_2\\600=c_2$$so our final solution is $$y(t)=600e^{0.15t}$$
anonymous
  • anonymous
so for \(t=5\) we have:$$y(5)=600e^{5(0.15)}=600e^{0.75}=600e^{3/4}\approx 1270$$
anonymous
  • anonymous
all correct, but that is an awful lot of work if the growth "factor" is \(.15\) then it is \[A_0e^{.15t}\]
anonymous
  • anonymous
OH, yeah its just been like months since Ive done separation and I didn't recognize it. thank you all
anonymous
  • anonymous
this is a pre calculus problem, requires only knowing about continuous compounding, not difeq just thought you might remember it that way
anonymous
  • anonymous
ugh, yeah! I always end up doing waaay took
anonymous
  • anonymous
i assume "took" is some code for "too much damn work"

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