The given DE is;
dtdn=20kn(k−n)
a] By separation of variables we have;
n(k−n)1dtdn=20k1
Integrating both sides wrt t yields;
∫n(k−n)1dtdn dt=∫20k1dt⇒∫n(k−n)1 dn=20k1∫ dt⇒∫(kn1+k(k−n)1) dn=20k1∫ dt⇒∫kn1 dn+∫k(k−n)1 dn=20k1∫ dt⇒kln(n)−kln(k−n)=20kt+c, where c is an arbitrary constant⇒ln(n)−ln(k−n)=20t+ck,⇒ln(k−nn)=0.05t+ck⇒k−nn=c1e0.05t, where c1=eck⇒ne−0.05t=c1(k−n)⇒ne−0.05t+nc1=c1k⇒n=c1+e−0.05tc1k=1+c11e−0.05t200=1+c2e−0.05t200, where c2=c11 and k=200
Now, at t=0, n=50 thus;
50=1+c2200⇒1+c2=4⇒c2=3.
Thus, n=1+3e−0.05t200
b] Since the question says after 2022, then the population of bears that will be able to survive after 2022 is;
n(2023)=1+3e−0.05×2023200=200 bears
Comments