Importance of Differential Equations:

Differential equations is a very important topic for both Mathematics as well as engineering students. Most of the problems that arise in engineering turns to Mathematical models involving Differential equations. So , a student has to learn how to form a differential equation from a physical problem , solve it by using suitable methods and interpret the results.

Before going to the actual topic, we learn about basics like what is Differential equation , types , order and degree.

I. What is Differential Equation ?

An equation involving differentials or differential coefficients is known as differential equation .

Ex :$$ \frac{dy}{dx}+ \cos x $$= 0

II. Types of Differential Equation :

Mainly Differential Equations are of two types. They are --

1. Ordinary Differential equation

2. Partial Differential equation

1. Ordinary Differential Equation 

An equation involving derivatives with respect to only one independent variable is called as Ordinary Differential Equation.

Ex: $$\frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + y $$= cosx

2. Partial Differential Equation

An equation involving derivatives with respect to more than one independent variable is called a Partial Differential Equation.

Ex: $$ \frac{\partial ^{2}u}{\partial x^{2}} + \frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}} $$ = 0

Initial Value Problem:

A problem having governing equation with initial conditions is called as Initial Value Problem.

III. Order of a Differential Equation:

The order of the highest ordered derivative occurring in a differential equation is called as Order of the Differential equation.

Ex: $$\frac{d^{2}y}{dx^{2}} + 3\frac{dy}{dx}+ y $$= 0

In the above example the order of the differential equation is 2 .

IV. Degree of a Differential Equation:

The Power of the highest ordered derivative free from radicals and fractions is called degree of differential equation .

Ex 1: $$ \left [ 1 + \left ( \frac{dy}{dx} \right )^{2} \right ]^{\frac{5}{2}} = \frac{d^{2}y}{dx^{2}} $$ ,

In the above example the Order is 2 and the degree is 2 .

Ex 2: $$ sin\left ( \frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx}\right ) $$= 0 , In this example the order is 2 and the degree is not defined .

Solution :

The relation between dependent and independent variable satisfying the given Differential equation and not containing derivatives is called as Solution of the Differential Equation .

Ex : y = c.e$$^{x}$$ as the solution of $$\frac{dy}{dx}$$ = y

General Solution :

The solution in which number of arbitrary constants is equal to the order of the Differential equation is called as General Solution of the given Differential equation.

Ex : y = $$A \cos X + B \sin X$$ is the General solution of $$\frac{d^{2}y}{dx^{2}}$$ + y = 0

Particular Solution :

The solution obtained from general solution by assigning particular values to arbitrary constants is called as Particular Solution .

If we take A = 1 and B = 2 in the above solution , we get particular solution of Y = $$\cos x + 2 \sin $$ x

Formation of Differential Equation:

Let $$ f (x_{1} ,x_{2},c_{1},c_{2} .........c_{n} ) $$ = 0 be any relation or where $$c_{1},c_{2} .........c_{n}$$ are arbitrary constants .

Differentiating the above equation with respect to ‘x’ n times successively and eliminating ‘n’ arbitrary constants from these (n+1) equations , we get a Differential equation of order 'n'.

Problem 1:

Obtain the Differential equation by eliminating arbitrary constant ‘a’ from$$ x^{2} + y^{2} +2 ax = C^{2}$$.

Solution :

Given Equation is$$ x^{2} + y^{2} +2ax $$ =$$C^{2}$$ ………………………..(1)

Differentiate equation (1) with respect ‘x’

2x+2y$$\frac{dy}{dx}$$ + 2a(1) = 0

2a = -$$\left ( 2x + 2y \frac{dy}{dx}\right )$$ ……………………….(2)

Substituting the equation (2) in (1) , we get

$$ X^{2} +y^{2} +\left (-\left (2x +2y\frac{dy}{dx}\right) \right)x$$ = $$C^{2}$$

$$X^{2} +y^{2} – 2x^{2} -2xy \frac{dy}{dx}$$ = $$C^{2}$$

$$y^{2} –X^{2} -2xy\frac{dy}{dx} $$ = $$C^{2}$$ , is an Ordinary Differential Equation of

Order 1 and degree 1.

Problem 2 :

Find the differential equations of all circles of radius ‘a’ and centre (h,k) , also find the order and degree of the differential equation .

Solution :

Equation of the circle with centre (h,k) and radius ‘a’ , h,k are parameters .

$$ (x-h)^{2} +(y-k)^{2}$$ = $$a^{2}$$ ……………………….(1)

Differentiate the above equation with respect to ‘x’.

2(x-h) + 2(y-k) y’ = 0 …………..(2)

$$\Rightarrow $$ (x-h) = -(y-k)y’ …………………(A)

Differentiate equation (2) with respect to ‘x’.

2(1) + 2(y-k) y’’ + 2 (y’)y’ = 0

$$ \Rightarrow$$ 2 + 2(y-k)y’’ + 2(y’)$$^{2}$$ = 0

(y-k) = -$$-\frac{1+(y')^{2}}{y^{''}}$$y’’ ……………………….(3)

Using (3) in (A) , we get

(x-h) = -$$ \left(\frac{1+(y’)^{2}}{y’’}\right) \times y’$$ ……………..(4)

Substitute equation (3) and (4) in equation (1) , we get

$$\left(\left(\frac{1+(y^{'})^2}{y^{''}}\right)y^{'}\right)^2+\left(- \left(\frac{1+(y^{'})^2}{y^{''}}\right)\right)^2$$=$$a^2$$

$$\frac{(1+y^{'})^{2}}{y^{''}}\times (y^{'})^{2}$$ + $$ \frac{(1+y^{'})^{2}}{y^{''}}$$ = $$a^{2}$$

$$\frac{(1+y^{'})^{2})^{2}}{y^{''}} \left [ (y^{'})^{2} + 1\right ] $$ = $$a^{2}$$

$$ \frac{[1+(y^{'})^2]^3}{(y^{''})^2}$$=$$a^2$$

$$ [1+(y^{'})^{2}]^3$$ = $$a^2(y^{''})^2$$ , is ordinary differential equation of Order --2 and degree --- 2