2MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30We willchoosethese constants so thatA= 1 andB−A= 0 i.e., we chooseA=B= 1. Simply put, theleft-hand side of the differential equation becomes1y(1−y)dydt=1y+11−ydydt=ddtln|y|−ln|1−y|.Upon integrating both sides of the differential equation, we find thatln|y|−ln|1−y|=r t+Cfor some constantC. This is a perfectly validimplicitsolution to the differential equation, but we want tosolve fory=y(t) to find anexplicitsolution. To this end, recall the following identity for the difference oftwo logarithms:lnA−lnB= lnAB.This means we havelny1−y=r t+C=⇒y1−y=C1ertwhereC1=±eCis a constant. For the initial condition, sett= 0:C1=y01−y0=⇒y(t)1−y(t)=y01−y0ertSolving fory=y(t), we find the solutiony(t) =y0y0+ (1−y0)e−rt=⇒P(t) =K y(t) =P0KP0+ (K−P0)e−rt.These are explicit solutions to the initial value problems.
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