# Applications of Differential Equations

We present examples where differential equations are widely applied to model natural phenomena, engineering systems, and many other situations.

## Application 1: Exponential Growth - Population

Let $$P(t)$$ be a quantity that increases with time $$t$$ and the rate of increase is proportional to the same quantity $$P$$ as follows $\frac{dP}{dt} = kP$ where $$\frac{dP}{dt}$$ is the first derivative of $$P$$, $$k > 0$$, and $$t$$ is the time. The solution to the above first-order differential equation is given by $P(t) = A e^{kt}$ where $$A$$ is a constant not equal to 0. If $$P = P_0$$ at $$t = 0$$, then $P_0 = A e^0$ which gives $$A = P_0$$. The final form of the solution is given by $P(t) = P_0 e^{kt}$ Assuming $$P_0$$ is positive and since $$k$$ is positive, $$P(t)$$ is an increasing exponential. $$\frac{dP}{dt} = kP$$ is also called an exponential growth model.

## Application 2: Exponential Decay - Radioactive Material

Let $$M(t)$$ be the amount of a product that decreases with time $$t$$ and the rate of decrease is proportional to the amount $$M$$ as follows $\frac{dM}{dt} = -kM$ where $$\frac{dM}{dt}$$ is the first derivative of $$M$$, $$k > 0$$, and $$t$$ is the time. Solve the above first-order differential equation to obtain $M(t) = A e^{-kt}$ where $$A$$ is a non-zero constant. If we assume that $$M = M_0$$ at $$t = 0$$, then $M_0 = A e^0$ which gives $$A = M_0$$. The solution may be written as follows $M(t) = M_0 e^{-kt}$ Assuming $$M_0$$ is positive and since $$k$$ is positive, $$M(t)$$ is a decreasing exponential. $$\frac{dM}{dt} = -kM$$ is also called an exponential decay model.

## Application 3: Falling Object

An object is dropped from a height at time $$t = 0$$. If $$h(t)$$ is the height of the object at time $$t$$, $$a(t)$$ the acceleration, and $$v(t)$$ the velocity. The relationships between $$a$$, $$v$$, and $$h$$ are as follows: $a(t) = \frac{dv}{dt}, \quad v(t) = \frac{dh}{dt}.$ For a falling object, $$a(t)$$ is constant and is equal to $$g = -9.8 \, \text{m/s}^2$$. Combining the above differential equations, we can easily deduce the following equation $\frac{d^2h}{dt^2} = g$ Integrate both sides of the above equation to obtain $\frac{dh}{dt} = gt + v_0$ Integrate one more time to obtain $h(t) = \frac{1}{2}gt^2 + v_0t + h_0$ The above equation describes the height of a falling object, from an initial height $$h_0$$ at an initial velocity $$v_0$$, as a function of time.

## Application 4: Newton's Law of Cooling

It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature $$T$$ of the object and the temperature $$T_e$$ of the environment surrounding the object. $\frac{dT}{dt} = -k(T - T_e)$ Let $$x = T - T_e$$ so that $$\frac{dx}{dt} = \frac{dT}{dt}$$. Using the above change of variable, the above differential equation becomes $\frac{dx}{dt} = -kx$ The solution to the above differential equation is given by $x = A e^{-kt}$ substitute $$x$$ by $$T - T_e$$ $T - T_e = A e^{-kt}$ Assume that at $$t = 0$$ the temperature $$T = T_0$$ $T_0 - T_e = A e^0$ which gives $$A = T_0 - T_e$$
The final expression for $$T(t)$$ is given by $T(t) = T_e + (T_0 - T_e)e^{-kt}$ This last expression shows how the temperature $$T$$ of the object changes with time.

## Application 5: RL Circuit

Let us consider the RL (resistor R and inductor L) circuit shown above. At $$t = 0$$, the switch is closed, and current passes through the circuit. Electricity laws state that the voltage across a resistor of resistance $$R$$ is equal to $$R i$$, and the voltage across an inductor $$L$$ is given by $$L \frac{di}{dt}$$ (where $$i$$ is the current). Another law gives an equation relating all voltages in the above circuit as follows: $L \frac{di}{dt} + Ri = E, \quad \text{where } E \text{ is a constant voltage}.$ Let us solve the above differential equation, which may be written as follows: $L \frac{\frac{di}{dt}}{E - Ri} = 1$ This can be written as: $-\frac{L}{R} \frac{-Rdi}{E - Ri} = dt$ Integrate both sides: $-\frac{L}{R} \ln(E - Ri) = t + c, \quad c \text{ constant of integration}.$ Find constant $$c$$ by setting $$i = 0$$ at $$t = 0$$ (when the switch is closed), which gives: $c = -\frac{L}{R} \ln(E)$ Substitute $$c$$ in the solution: $-\frac{L}{R} \ln(E - Ri) = t - \frac{L}{R} \ln(E)$ This may be written as: $\frac{L}{R} \ln\left(\frac{E}{E - Ri}\right) = t$ Change into exponential form: $\frac{E}{E - Ri} = e^{t(R/L)}$ Solve for $$i$$ to obtain: $i = \frac{E}{R} \left(1 - e^{-\frac{Rt}{L}}\right)$ The starting model for the circuit is a differential equation which, when solved, gives an expression of the current in the circuit as a function of time.