CBSE Class 12 Mathematics Differential Equations Case Studies

Differential Equations

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: \( \frac{dT}{dt} \propto (T - 70) \), where 70°F is the room temperature and \( T \) is the temperature of the object at time \( t \). Substituting the two different observations of \( T \) and \( t \) made, in the solution of the differential equation \( \frac{dT}{dt} = k(T - 70) \) where \( k \) is a constant of proportion, time of death is calculated.

Question. State the degree of the above given differential equation.
(a) 0
(b) 1
(c) 2
(d) Not defined
Answer: (b) We have differential equation \( \frac{dT}{dt} = k(T - 70) \) with degree 1.

Question. Which method of solving a differential equation helped in calculation of the time of death?
(a) Variable separable method
(b) Solving Homogeneous differential equation
(c) Solving Linear differential equation
(d) all of the above
Answer: (a) To solve the differential equation \( \frac{dT}{dt} = k(T - 70) \), Variable-Separable method is useful.

Question. If the temperature was measured 2 hours after 11.30 pm, will the time of death change? (Yes/No)
(a) Yes
(b) No
(c) Cannot be determined
(d) None of these
Answer: (b) No, the time of death would not change.

Question. The solution of the differential equation \( \frac{dT}{dt} = k(T - 70) \) is given by,
(a) \( \log |T - 70| = kt + C \)
(b) \( \log |T - 70| = \log |kt| + C \)
(c) \( T - 70 = kt + C \)
(d) \( T - 70 = ktC \)
Answer: (a) We have, \( \frac{dT}{dt} = k(T - 70) \Rightarrow \int \frac{dT}{T - 70} = \int kdt \Rightarrow \log |T - 70| = kt + C \).

Question. If \( t = 0 \) when \( T \) is 72, then the value of \( C \) is
(a) –2
(b) 0
(c) 2
(d) \( \log 2 \)
Answer: (d) Given \( t = 0 \) when \( T = 72 \). Now, \( \log |T - 70| = kt + C \Rightarrow \log |72 - 70| = k \times 0 + C \Rightarrow \log 2 = C \).

Polio drops are delivered to 50K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of \( 2^{nd} \) week half the children have been given the polio drops. How many will have been given the drops by the end of \( 3^{rd} \) week can be estimated using the solution to the differential equation \( \frac{dy}{dx} = k(50 - y) \) where \( x \) denotes the number of weeks and \( y \) the number of children who have been given the drops.

Question. The order of the above given differential equation is
(a) 1
(b) 2
(c) 3
(d) none of these
Answer: (a) Given differential equation \( \frac{dy}{dx} = k(50 - y) \) has order 1.

Question. Which method of solving a differential equation can be used to solve \( \frac{dy}{dx} = k(50 - y) \)?
(a) Variable separable method
(b) Solving Homogeneous differential equation
(c) Solving Linear differential equation
(d) all of the above
Answer: (a) Variable Separable method can be used to solve differential equation \( \frac{dy}{dx} = k(50 - y) \).

Question. The solution of the differential equation \( \frac{dy}{dx} = k(50 - y) \) is given by,
(a) \( \log |50 - y| = kx + C \)
(b) \( -\log |50 - y| = kx + C \)
(c) \( \log |50 - y| = \log |kx| + C \)
(d) \( 50 - y = kx + C \)
Answer: (b) We have, \( \frac{dy}{dx} = k(50 - y) \Rightarrow \int \frac{dy}{50 - y} = \int kdx \Rightarrow -\log |50 - y| = kx + C \).

Question. The value of \( C \) in the particular solution given that \( y(0) = 0 \) and \( k = 0.049 \) is.
(a) \( \log 50 \)
(b) \( \log \frac{1}{50} \)
(c) 50
(d) –50
Answer: (b) Given \( y(0) = 0 \) and \( k = 0.049 \). \( -\log |50 - y| = kx + C \Rightarrow -\log |50 - 0| = 0.049 \times 0 + C \Rightarrow -\log 50 = C \Rightarrow C = \log \frac{1}{50} \).

Question. Which of the following solutions may be used to find the number of children who have been given the polio drops?
(a) \( y = 50 - e^{kx} \)
(b) \( y = 50 - e^{-kx} \)
(c) \( y = 50(1 - e^{-kx}) \)
(d) \( y = 50(e^{kx} - 1) \)
Answer: (c) We have, \( -\log |50 - y| = kx + \log \frac{1}{50} \Rightarrow -kx = \log |50 - y| + \log \frac{1}{50} \Rightarrow -kx = \log \frac{50 - y}{50} \Rightarrow e^{-kx} = \frac{50 - y}{50} = 1 - \frac{y}{50} \Rightarrow \frac{y}{50} = 1 - e^{-kx} \Rightarrow y = 50(1 - e^{-kx}) \). This is the required solution to find the number of children who have been given the polio drops.