Consider the initial value problem dy/dx = y^2 - 5y + 6 y(0) = 1 where y represe
ID: 2859144 • 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 Vector -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 Vector -Infinity y(x)Explanation / Answer
given dy/dx =y2-5y+6
dy/(y2-5y +6)=dx
dy/(y2-2y-3y +6)=dx
dy/(y(y-2)-3(y -2))=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
given y(0)=0
ln((1-3)/(1-2))=0 +c
ln2=c
ln((y-3)/(y-2))=x +ln2
((y-3)/(y-2))=ex +ln2
((y-3)/(y-2))=exeln2
((y-3)/(y-2))=2ex
1-(1/(y-2))=2ex
(1/(y-2))=1-2ex
(y-2)=(1/(1-2ex))
y =2+(1/(1-2ex))
y =((2-4ex+1)/(1-2ex))
y =((3-4ex)/(1-2ex)) is the solution
b)limx-> y(x)
=limx-> ((3-4ex)/(1-2ex))
=limx-> (ex(3e-x-4)/ex(1e-x-2))
=limx-> ((3e-x-4)/(1e-x-2))
=(0-4)/(0-2)
=4/2
=2
population approaches 2 thousand
c)limx->- y(x)
=limx->- ((3-4ex)/(1-2ex))
= ((3-0)/(1-0))
=3
population approaches 3 thousand