Consider the initial value problem dy/dx = y^2 - 5y + 6 y (0) = 1 where y repres
ID: 2876532 • Letter: C
Question
Consider the initial value problem dy/dx = y^2 - 5y + 6 y (0) = 1 where y represents the population of a species (in thousands), and x represents time (in years). Solve this initial value problem to find the population as a function of time. Write your solution explicitly. What number does the population approach in the long term? (i.e., find an equilibrium solution by computing lim_x rightarrow infinity y(x)) What number would the population approach if one could, theoretically, go back in time? (i.e., find an equilibrium solution by computing lim_x rightarrow -infinity y(x))Explanation / Answer
5)given dy/dx =y2-5y +6
a) seperate the variables
dy/(y2-5y +6)=dx
dy/((y-2)(y-3))=dx
(1/(y-3) -1/(y-2))dy =dx
integrate on both sides
(1/(y-3) -1/(y-2))dy =dx
ln(y-3)-ln(y-2)=x+c
ln((y-3)/(y-2))=x+c
((y-3)/(y-2))=ex+c
((y-3)/(y-2))=Cex
given y(0)=1
((1-3)/(1-2))=Ce0
2=C*1
C=2
((y-3)/(y-2))=2ex
(y-3)=(y-2)2ex
y-3=2yex-4ex
y-2yex=3-4ex
y(1-2ex)=(3-4ex)
y=(3-4ex)/(1-2ex)
y=(4ex-3)/(2ex-1)
b)
limx->y(x)
=limx->(4ex-3)/(2ex-1)
=limx->ex(4-3e-x)/ex(2-e-x)
=limx->(4-3e-x)/(2-e-x)
=(4-0)/(2-0)
=4/2
=2
in long term population approach 2 thousand
c)
limx->-y(x)
=limx->-(4ex-3)/(2ex-1)
=(0-3)/(0-1)
=3
when gone back in time population approach 3 thousand