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: `(dT)/(dt) prop (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 `(dT)/(dt)`=k(T-70) where k is a constant of proportion, time of death is calculated.
Which method of solving a differential equation helped in calculation of the time of death?

Options

A. Variable separable method
B. Solving Homogeneous differential equation
C. Solving Linear differential equation
D. all of the above

Answer

To solve the problem of estimating the time of death of the cat using Newton's law of cooling, we will follow these steps:

### Step 1: Understand the differential equation
The differential equation given is:
\[ \frac{dT}{dt} = k(T - 70) \]
where \( T \) is the temperature of the cat at time \( t \), \( 70^\circ F \) is the room temperature, and \( k \) is a constant.

### Step 2: Separate the variables
We can rearrange the equation to separate the variables:
\[ \frac{dT}{T - 70} = k \, dt \]

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dT}{T - 70} = \int k \, dt \]
This gives us:
\[ \ln |T - 70| = kt + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for \( T \)
Exponentiating both sides to eliminate the logarithm, we have:
\[ |T - 70| = e^{kt + C} \]
Let \( C' = e^C \), then:
\[ T - 70 = C'e^{kt} \]
Thus,
\[ T = 70 + C'e^{kt} \]

### Step 5: Use initial conditions to find \( C' \) and \( k \)
We have two temperature readings:
1. At \( t = 0 \) (11:30 PM), \( T = 94.6^\circ F \)
2. At \( t = 1 \) (12:30 AM), \( T = 93.4^\circ F \)

Using the first observation:
\[ 94.6 = 70 + C'e^{k \cdot 0} \implies C' = 24.6 \]
So,
\[ T = 70 + 24.6e^{kt} \]

Using the second observation:
\[ 93.4 = 70 + 24.6e^{k \cdot 1} \]
This simplifies to:
\[ 23.4 = 24.6e^{k} \implies e^{k} = \frac{23.4}{24.6} \]

### Step 6: Calculate \( k \)
Taking the natural logarithm:
\[ k = \ln\left(\frac{23.4}{24.6}\right) \]

### Step 7: Estimate the time of death
To find the time of death, we need to determine when \( T = 98.6^\circ F \):
\[ 98.6 = 70 + 24.6e^{kt_d} \]
Solving for \( t_d \) gives us:
\[ 28.6 = 24.6e^{kt_d} \implies e^{kt_d} = \frac{28.6}{24.6} \]
Taking the natural logarithm:
\[ kt_d = \ln\left(\frac{28.6}{24.6}\right) \implies t_d = \frac{1}{k} \ln\left(\frac{28.6}{24.6}\right) \]

### Conclusion
The method used to solve the differential equation is the **variable separable method**.

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