Introduction to Differential Equation
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.
An equation involving differentials or differential coefficients is known as differential equation .
Ex : \frac{dy}{dx}+ \cos x
Mainly Differential Equations are of two types. They are --
1. Ordinary Differential equation
2. Partial 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
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}}
A problem having governing equation with initial conditions is called as Initial Value Problem.
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
In the above example the order of the differential equation is 2 .
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 )
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}
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
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
Formation of Differential Equation:
Let f (x_{1} ,x_{2},c_{1},c_{2} .........c_{n} )
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
Differentiate equation (1) with respect ‘x’
2x+2y\frac{dy}{dx}
2a = -\left ( 2x + 2y \frac{dy}{dx}\right )
Substituting the equation (2) in (1) , we get
X^{2} +y^{2} +\left (-\left (2x +2y\frac{dy}{dx}\right) \right)x
X^{2} +y^{2} – 2x^{2} -2xy \frac{dy}{dx}
y^{2} –X^{2} -2xy\frac{dy}{dx}
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}
Differentiate the above equation with respect to ‘x’.
2(x-h) + 2(y-k) y’ = 0 …………..(2)
\Rightarrow
Differentiate equation (2) with respect to ‘x’.
2(1) + 2(y-k) y’’ + 2 (y’)y’ = 0
\Rightarrow
(y-k) = --\frac{1+(y')^{2}}{y^{''}}
Using (3) in (A) , we get
(x-h) = - \left(\frac{1+(y’)^{2}}{y’’}\right) \times y’
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
\frac{(1+y^{'})^{2}}{y^{''}}\times (y^{'})^{2}
\frac{(1+y^{'})^{2})^{2}}{y^{''}} \left [ (y^{'})^{2} + 1\right ]
\frac{[1+(y^{'})^2]^3}{(y^{''})^2}
[1+(y^{'})^{2}]^3
1 Doubts's
I did not understand the concept, more detail plz