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

rl circuit for application 5
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

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