Ordinary Differential Equations Questions and Answers – Law of Natural Growth and Decay

This set of Ordinary Differential Equations Interview Questions and Answers for Experienced people focuses on “Law of Natural Growth and Decay”.

1. The population of the Mysore city of a country was 30 lakhs in the year 2000 and 60 lakhs in the year 2010. What is the estimated population in the year 1990 with the help of the population model?
a) 18 lakhs
b) 15 lakhs
c) 20 lakhs
d) 12 lakhs
View Answer

Answer: b
Explanation: Here rate of change of population is proportional to the population present at that instant of time i.e \(\frac{dp}{dt} ∝ P\) i.e \(\frac{dp}{dt}=kP\), k is a constant of proportionality.
solving using the variable separable integral form we get \(\int \frac{1}{P} \,dP = \int K \,dt + a\) log P=kt+c –> ekt+a = P or P=cekt…where ea is a constant let the year 2000 be the initial reference year –> P(0)=30=cek(0) = c i.e c = 30 & P(10) = cek10
=30ek10 = 60 –> log 60 = log 30 + 10k –> log 63 =log 2 = 10k –> k = 0.1*0.693 = 0.0693
to find P(-10) = cekt = 30*e-0.693=15 lakhs.

2. Urine culture test is being conducted in the pathology lab from the sample of a patient to prescribe a suitable drug based on bacterial growth. There were 3000 bacteria in the initial sample and it had increased to 3400 in the sample collected after one hour. Based on the exponential growth how long will it take for the bacterial population to reach 10000?
a) 10 hrs
b) 8.5 hrs
c) 9.6 hrs
d) 3.6 hrs
View Answer

Answer: c
Explanation: As we know according to law of exponential growth, the bacterial population model is
B(t)=cekt at t=0 B(0)=ce0=3000……given –> c=3000 and B(1)=3400=cek=3000*ek
ek=\(\frac{17}{15}\) taking log, k=log 17 – log 15 = 0.125, to find t when B(t)=10000 i.e
10000=3000*ekt = 3000*e0.125t –> 3.33=e0.125t –> log 3.33 = 0.125t –> t=9.62 hrs.

3. Uranium disintegrates at a rate proportional to the amount present at any time instant. If a and b grams of uranium are present at times t’ & t’’ respectively. What is the expression for the half life of uranium?
a) \(\frac{(t”+t’)log2}{log⁡(\frac{b}{a})} \)
b) \(\frac{(t”-2t’)log2}{2log(\frac{a}{b})} \)
c) \(\frac{(t”+t’)}{2log(\frac{a}{b})} \)
d) \(\frac{(t”-t’)log2}{log⁡(\frac{b}{a})} \)
View Answer

Answer: d
Explanation: Let P grams of uranium be present at any time t. According to law of decay, \(\frac{dp}{dt} = -kP…k\) is a constant its solution is P=ce-kt given that at t=t’ P=a
a=ce-kt…..(1) and b=ce-kt….(2) dividing (1) & (2) we get \(\frac{a}{b}=e^{k(t^{”}-t’)}\) taking log on both side \(log(\frac{a}{b})\) = k(t’’-t’)…(3) at t=0 ‘c’ is the initial mass of uranium present in the sample thus at half life i.e t=T ,P=c/2 P = P=ce-kt becomes \(\frac{c}{2}\) = ce-kT
log 2=kT substituting the value of k from (3) we get \(T = \frac{(t”-t’)log2}{log⁡(\frac{b}{a})} \) is the equation for half life period.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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