Question

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.)

Answer 1

\[\frac{ dy }{ y }=0.15 dt\]
integrate both sides

Answer 2

could you explain further please?

Answer 3

What are you confused about maria? :)
Are you familiar with `Separation of Variables`?

Answer 4

@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 \(c_2=e^{c_1}\)

Answer 5

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 $$y(t)=600e^{0.15t}$$

Answer 6

so for \(t=5\) we have:$$y(5)=600e^{5(0.15)}=600e^{0.75}=600e^{3/4}\approx 1270$$

Answer 7

all correct, but that is an awful lot of work
if the growth "factor" is \(.15\) then it is
\[A_0e^{.15t}\]

Answer 8

OH, yeah its just been like months since Ive done separation and I didn't recognize it. thank you all

Answer 9

this is a pre calculus problem, requires only knowing about continuous compounding, not difeq
just thought you might remember it that way

Answer 10

ugh, yeah! I always end up doing waaay took

Answer 11

i assume "took" is some code for "too much damn work"