Question

Write a detail note on applications of Differential Equation

Answer 1

Differential equations are important as they are used in calculation of the rates of change of various systems

E.g.

1.Exponential population growth

Let P(t) be a quantity increasing with time, t and the rate of increase is proportional to P

I.e.

$\frac{dP}{dt}=kP$

Solution to this differential equation is given by

$P(t)=A e^{kt}$

2. Exponential Decay- Radioactive Material

Let M(t) be the amount of a product decreasing with time, t and decreasing rate is proportional to M

I.e.

$\frac{dM}{dt}=-kM$

Where, $\frac{dM}{dt}$ is the first derivative of M

Solving this we get,

$M(t)=A e^{-kt}$

M(t) = A e^{- k t}

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

$a(t)=\frac{dv}{dt}, v(t)=\frac{dh}{dt}$

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.

$\frac{dT}{dt}=-k(T-Te)$

Let x= T - Te hence, $\frac{dx}{dt}={dT}{dt}$

Using this, the above formula becomes

$\frac{dx}{td}=-kx$

Hence we have, $x=A e^{-kt}$

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.

5. RL Circuit

Consider an RL circuit

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 $L\frac{di}{dt}$ (I is current).