17. Applications of Differential Equations

a. Exponential Growth and Decay

The most well-known example of a differential equation is \(\dfrac{dy}{dt}=\pm k y\), in which the rate of change of a quantity is proportional to the amount of the quantity. The solutions describe exponential growth when the coefficient is positive and exponential decay when the coefficient is negative. These were covered briefly in the previous chapter. We now solve these equations in general and use them in applications. Since growth and decay usually occur over time, we usually take the independent variable to be \(t\).

Find the general solution of: \[\dfrac{dy}{dt}=ky \qquad \text{where} \qquad k>0\] and then find the specific solution satisfying the initial condition: \[ y(0)=y_o \]

We solve the differential equation by separating variables and integrating: \[ \int \dfrac{dy}{y}=\int k\,dt \qquad \Longrightarrow \qquad \ln|y|=kt+C \] To solve for \(y\), we exponentiate and then remove the absolute values: \[ |y|=e^C e^{kt} \qquad \Longrightarrow \qquad y=\pm e^C e^{kt} \] To get a final form for the solution, we switch from an additive constant of integration \(C\), to a multiplicative constant of integration defined by \(A=\pm e^C\). So the general solution is: \[ y=Ae^{kt} \] The initial condition \(y(0)=y_o\) says \(y=y_o\) when \(t=0\). Plugging into the solution says \(y_o=Ae^{k0}=A\). So the specific solution is: \[ y=y_o e^{kt} \]

The original definition, \(A=\pm e^C\), says \(A\ne0\) since \(e^C\) is never zero. However, the exceptional case, \(A=0\), is also allowed since \(y=0\) is also a solution of \(\dfrac{dy}{dt}=ky\). This execptional solution was lost when we separated variables and divided by \(y\) to get \(\dfrac{dy}{y}=k\,dt\).

The bacteria in a culture reproduce at a rate which is proportional to the amount of bacteria present. Consequently, the number of bacteria \(N(t)\) satisfies \[\dfrac{dN}{dt}=kN \qquad \text{where} \qquad k>0\] If there are initially \(N(0)=4\times10^6\) bacteria and after \(3\) hours there are \(N(3)=7\times10^6\) bacteria, how many bacteria will there be after \(5\) hours?

At present, we do not know the constant \(k\). However, we do know the solution is: \[ N=N_o e^{kt} \qquad \text{where} \qquad N_o=4\times10^6 \] This is exponential growth. We now use the second data point, \(N(3)=7\times10^6\), to find the growth constant \[ N(3)=4\times10^6e^{k3}=7\times10^6 \qquad \Longrightarrow \qquad e^{k3}=\dfrac{7}{4} \qquad \Longrightarrow \qquad k=\dfrac{1}{3}\ln\dfrac{7}{4} \] So the solution is \[ N(t)=4\times10^6e^{(t/3)\ln(7/4)} \] Finally, we compute the number of bacteria at \(t=5\): \[ N(5)=4\times10^6e^{(5/3)\ln(7/4)} \approx1.0165\times10^{7}\]

Find the general solution of: \[\dfrac{dy}{dt}=-ky \qquad \text{where} \qquad k>0\] and then find the specific solution satisfying the initial condition: \[ y(0)=y_o \]

The solution is the same as that for exponential growth, except \(k\) is replaced everywhere by \(-k\). So the general solution is: \[ y=Ae^{-kt} \] and the specific solution satisfying the initial condition is: \[ y=y_o e^{-kt} \]

Notice that the solution can be rewritten as: \[ Q(t)=40e^{-(t/1590)\ln2}=40\left(e^{-\ln2}\right)^{t/1590}=40\left(\dfrac{1}{2}\right)^{t/1590} \] This can be interpreted as saying every time the time increases by \(1590\), the quantity gets multiplied by \(\dfrac{1}{2}\).

In general, if the initial amount is \(Q_o\) and the half-life is \(\tau\), then the formula is \(Q(t)=Q_o\left(\dfrac{1}{2}\right)^{t/\tau}\).

Similarly, if the initial amount (of bacteria) is \(N_o\) and the doubling time is \(\tau\), then the formula is \(N(t)=N_o2^{t/\tau}\).

The exponential growth and decay equations are examples of autonomous differential equations which involve the dependent function and its derivatives but have no explicit dependence on the independent variable. A first order autonomous differential equation would have the form: \[ \dfrac{dy}{dt}=F(y) \qquad \text{for some function} \qquad F(y) \] These equations can usually be solved by separation of variables. The example on this page and most of the rest of this chapter are autonomous differential equations.