Total students =100 At t=0 (zero day), infected student =2 Let at t=t day infected student =x ∴ At t=t day non infected student =(100−x) Rate of infection =
dx
dt
Given,
dx
dt
∝x(100−x) ⇒∫
dx
x(100−x)
=∫kdt ⇒
1
100
∫
100−x+x
x(100−x)
dx=kt+c ⇒
1
100
∫(
1
x
+
1
100−x
)dx=kt+c ⇒
1
100
[lnx−ln(100−x)]=kt+c
⇒1∕100lnx∕100−x=kt+c Given, At, t=0,x=2 ∴1∕100ln2∕98=c Putting value of c in equation (1), we get 1∕100lnx∕100−x=kt+1∕100ln2∕98 ⇒1∕100lnx∕100−x−1∕100ln2∕98=kt ⇒1∕100lnx×98∕2(100−x)=kt Given, At t=4,x=30 ∴1∕100ln30×98∕2(70)=k×4 ⇒k=1∕400ln21 ∴1∕100lnx×98∕2(100−x)=t×1∕400×ln21