# 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.

## More References and Links

Differential Equations

Solve Differential Equations Using Laplace Transform