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:
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:
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:
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 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:
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 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:
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.