Differential Equations

To understand Differential equations, let us consider this simple example. Have you ever thought why a hot cup of coffee cools down when kept under normal conditions? According to Newton, cooling of a hot body is proportional to the temperature difference between its own temperature T and the temperature Ts of its surrounding. This statement in terms of mathematics can be written as:

\frac{dT}{dt}

∝

(T – Ts)

…………(1)

Introducing a proportionality constant $k$ , the above equation can be written as:

\frac{dT}{dt}

k(T – Ts)

…………(2)

Here, T is the temperature of the body,

t is the time,

Ts is temperature of the surrounding,

\frac{dT}{dt}

is the rate of cooling of the body

Now the question arises how to calculate the time taken for the coffee to cool down or what is the significance of the above equation? The above equation is generally known as a differential equation of degree 1.

Differential equations are mathematical equations that relate some function with its derivatives. The functions generally represent any physical quantity and its derivatives represent the rate of change of the physical quantity. Differential equations relate two quantities in an equation which can be used in different ways as and when required. Every activity in this world can be related and unfolded by using differential equations and hence it plays a significant role in providing mathematical models for complex situations in science. Differential equations have been an important part of the modern developments in science and technology.

Application of differential equations:

Differential equations describe various exponential growths and decays.
They are also used to describe the change in investment return over time.
They are used in the field of medicines for modelling cancer growth or spread of a disease in the body.
Movement of electricity can also be described with the help of differential equations.
They help economists in finding the optimum investment strategies.
The motion of waves or a pendulum can also be described using differential equations.

Illustration 1: Verify that the function y = $e^{-3x}$ is a solution to the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} – 6y$ = 0.

Solution: The function given is y = $e^{-3x}$ . We differentiate both the sides of the equation with respect to x,

\frac{dy}{dx}

- 3 e^{-3x}

…………(1)

Now we again differentiate the above equation with respect to x,

\frac{d^2y}{dx^2}

9 e^{-3x}

…………(2)

We substitute the values of $\frac{dy}{dx}, \frac{d^2y}{dx^2}$ and y in the differential equation given in the question,

On left hand side we get, LHS = $9e^{-3x}+(-3e^{-3x})– 6e^{-3x}$

= $9e^{-3x} –9e^{-3x}$ = 0 (which is equal to RHS)

Therefore the given function is a solution to the given differential equation.

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Differential equations: A quick revision

Differential equations