Differential Equations Part 1

Let x be the number of bacteria at any time t.
Given that, ‌

dx

dt

∝x‌‌(∵ Rate of growth =‌

dx

dt

)
⇒‌‌‌

dx

dt

=λx⇒‌

dx

x

=λdt
After integrating it, we get
log‌x‌‌=λt+C . . . (i)
=0,x‌‌=1000‌ which gives ‌
log‌1000=0+C⇒C=log‌1000
Given, when t=0,x= log‌1000
log‌x−log‌1000=λt‌ or ‌log(‌

x

1000

)=λt . . . (ii)
Given that in 2h, number of bacteria increased by 20% i.e. when t=2h,x=1200 Put, t=2 and x=1200 in Eq. (ii), log(‌

1200

1000

)=2λ gives, λ=‌

1

2

‌log(‌

6

5

)
Again, from Eq. (ii),
log(‌

x

1000

)‌‌=‌

1

2

‌log(‌

6

5

)
or ‌

x

1000

‌‌=e‌

t

2

‌log(‌

6

5

). . . (iii)
Given, x=2000 at t=k∕loge(6∕5),
put in Eq. (iii),
‌

2000

1000

‌‌=e‌

k

2

‌log(‌

6

5

)∕log(‌

6

5

)
2‌‌=ek∕2or‌log‌2=k∕2
k∕log‌2‌‌=2
∴‌‌(k∕loge2)2‌‌=(2)2=4