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

d P / d t = k P

where d p / d t 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^{k t}

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^{k t}

Assuming P_{0} is positive and since k is positive, P(t) is an increasing exponential. d P / d t = k P 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

d M / d t = - k M

where d M / d t 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^{- k t}

where A is non zero constant.

It 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^{- k t}

Assuming M_{0} is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M 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) = dv / dt , v(t) = dh / dt.

For a falling object, a(t) is constant and is equal to g = -9.8 m/s.

Combining the above differential equations, we can easily deduce the follwoing equation

d^{ 2}h / dt^{ 2} = g

Integrate both sides of the above equation to obtain

dh / dt = g t + v_{0}

Integrate one more time to obtain

h(t) = (1/2) g t^{2} + v_{0} t + 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 Te of the environment surrounding the object.

d T / d t = - k (T - Te)

Let x = T - Te so that dx / dt = dT / dt

Using the above change of variable, the above differential equation becomes

d x / d t = - k x

The solution to the above differential equation is given by

x = A e ^{ - k t}

substitute x by T - Te

T - Te = A e ^{ - k t}

Assume that at t = 0 the temperature T = To

To - Te = A e ^{0}

which gives A = To - Te

The final expression for T(t) i given by

T(t) = Te + (To - Te)e ^{ - k t}

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 di/dt (i is the current). Another law gives an equation relating all voltages in the above circuit as follows:

L di/dt + Ri = E , where E is a constant voltage.

Let us solve the above differential equation which may be written as follows

L [ di / dt ] / [E - R i] = 1

which may be written as

- (L / R) [ - R d i ] / [E - Ri] = dt

Integrate both sides

- (L / R) ln(E - R i) = t + c , c constant of integration.

Find constant c by setting i = 0 at t = 0 (when switch is closed) which gives

c = (-L / R) ln(E)

Substitute c in the solution

- (L / R) ln(E - R i) = t + (-L/R) ln (E)

which may be written

(L/R) ln (E)- (L / R) ln(E - R i) = t

ln[E/(E - Ri)] = t(R/L)

Change into exponential form

[E/(E - Ri)] = e^{t(R/L)}

Solve for i to obtain

i = (E/R) (1-e^{-Rt/L})

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.