91Ó°ÊÓ

A differential equation relates a function with its derivatives. Imagine you have a quantity that changes over time, like the number of cells in a culture. A differential equation can describe how fast this number changes. In this exercise, the rate of growth of the cell culture is given by \(\frac{dN}{dt} = kN\), where \(N(t)\) is the number of cells at time \(t\), and \(k\) is a constant. This specific type is called a 'first-order' differential equation because it involves the first derivative of \(N\).
To solve it, we separate variables to get \(\frac{dN}{N} = k \, dt\), and then integrate both sides. Integration gives us \(\text{ln}(N) = kt + C\), where \(\text{ln}\) is the natural logarithm, \(k\) is our proportionality constant, and \(C\) is an integration constant.
Understanding differential equations is key to describing systems that change over time. In real life, they are used in fields ranging from biology to economics and engineering.

Initial conditions provide necessary information to find a particular solution to a differential equation. They tell us the state of the system at the beginning. In our problem, we know that at \(t = 0\), the cell population \(N \) is 1 million. This information allows us to determine the constant \(A\) in our general solution \(N = Ae^{kt}\).
When we substitute these initial conditions into the general solution, we get \(1 = A e^{k \times 0}\), simplifying to \(A = 1\). So, the particular solution becomes \(N = e^{kt}\).
Another initial condition given is that after 10 hours, the population is 9 million cells. By substituting \(t = 10\) and \(N = 9\), we get \(9 = e^{10k}\). Solving for \(k\) gives us the specific growth rate, which we can then use to predict future values.

The proportionality constant \(k\) determines the speed of growth in our differential equation. It links the rate of change to the current amount. In our case, it specifies how quickly the cell culture grows.
From the initial conditions that after 10 hours the population grows from 1 million to 9 million, we can determine \(\frac{dN}{dt} = kN\) by substituting into \(9 = e^{10k}\). Solving \(k = \frac{\text{ln}(9)}{10}\) gives \(k \), which is critical for finding the number of cells at any future time \(t\).
After computing \(k\), we use it in the equation \(N = e^{kt}\) to predict the culture's size at different times. For example, at 15 hours, it leads us to \(N(15) = e^{k \times 15} = 27\) million cells.
Understanding the proportionality constant helps in predicting how systems evolve over time.