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Engineering Mathematics – I (MAT-11)
By Dr. K.S. Chandrashekhar Professor & Head, Dept. of Mathematics, N.I.E., Mysore-
Lesson 1 Differential Equations
Session – 1
Introduction
Many problems in all branches of science and engineering when analysed for putting in a mathematical form assumes the form of a differential equation.
An engineer or an applied mathematician will be mostly interested in obtaining a solution for the associated equation without bothering much on the rigorous aspects. Accordingly the study of differential equations at various levels is focused on the methods of solving the equations.
Preliminaries
Ordinary Differential Equation (O.D.E)
If y = f ( x ) is an unknown function, an equation which involves atleast one derivative of y, w.r.t. x is called an ordinary differential equation which in future will be simply referred to as Differential Equation (D.E).
The order of D.E is the order of the highest derivative present in the equation and the degree of the D.E. is the degree of the highest order derivative after clearing the fractional powers.
Finding y as a function of x explicitly [ y = f ( x )] or a relationship in x and y satisfying the
D.E. [ f ( x, y )= c ] constitutes the solution of the D.E.
Observe the following equations along with their order and degree.
[order = 1, degree = 1]
[order = 1, degree = 2]
General solution and particular solution
A solution of a D.E. is a relation between the dependent and independent variables
satisfying the given equation identically.
The general solution will involve arbitrary constants equal to the order of the D.E.
If the arbitrary constants present in the solution are evaluated by using a set of given
conditions then the solution so obtained is called a particular solution. In many physical problems these conditions can be formulated from the problem itself.
Note : Basic integration and integration methods are essential prerequisites for this chapter.
Solution of differential equations of first order and first degree
Recollecting the definition of the order and the degree of a D.E., a first order and first degree equation will be the form
We discuss mainly classified four types of differential equations of first order and first degree. They as are as follows :
- Variables separable equations
- Homogenous equations
- Exact equations
- (^) Linear equations
Variables separable Equations
If the given D.E. can be put in the form such that the coefficient of dx is a function of the variable x only and the coefficient of dy is a function of y only then the given equation is said to be in the separable form.
The modified form of such an equation will be,
P ( x ) dx + Q ( y ) dy = 0
Integrating we have
This is the general solution of the equation.
Example –
Solve :
Dividing throughout by ( y -1) ( e x^ + 4) we get,
Session – 2
Equations reducible to the variables separable form
Some differential equations can be reduced to the variables separable from by taking a suitable substitution. We present a few examples.
Example – 6
Solve : cos ( x + y + 1) dx – dy = 0
Example – 7
Example – 8
Rearranging the terms in the given equation we have,
Let t + 1 = l (3 t 1) + m
or t + 1 = ( 3 l ) t + ( l+m )
Thus (2) can be written as
Example – 9 :
The given equation becomes,
Let 2 t + 3 = l (3 t + 4) + m
or 2 t + 3 = (3 l ) t + (4 l + m )
Thus 2 t + 3 = 2/3. (3 t+ 4) + 1/
Example – 10
The given d.e. can be written as,
(Observe that the coefficient of dx and dy are homogeneous functions of degree 3)
The given equation can be written as,
( for animation)*
Example –
Example –
Example – 4
Example – 5
Example – 6
Session – 4
Equations reducible to the homogeneous form
Consider the differential equation in the form :
We first express the equation in respect of dy / dx and the procedure is narrated by taking
This condition implies that there are no common factors for the x and y terms in the numerator as well as in the denominator.
Put x = X + h, y = Y + k where h and k are constants to be chosen appropriately later.
As a consequence of these (1) becomes
Now, let us choose h and k such that :
Solving these equations we get the value for h and k.
Thus (2) now assumes the form
It is evident that (3) is a homogeneous equation in the variables X and Y. This equation can be solved by putting Y = VX as discussed already. Finally we substitute for X and Y where X = x h, Y = yk.
Example –
Thus (1) becomes
Example – 2
Example – 3
Equations reducible to the exact form
Sometimes the given differential equation which is not an exact equation can be
transformed into an exact equation by multiplying with some function (factor) known as the integrating factor (I.F.)
The procedure to find such a factor is as follows.
Suppose that, for the equation M dx + N dy = 0
Example – 4
Example – 5
EXERCISES:
Solve the following differential equations
- cos x ( ey^ + 1) dx + sin x e y^ dy = 0
- [4 x^3 y^2 +^ y^ cos ( xy )]^ dx+ [2 x^4 y+x^ cos ( xy )]^ dy =^0
ANSWERS:
Session – 6
Linear Equations
A differential equation of the form
…(1)
where P and Q are functions of x only is called a linear equation in ‘ y ’.
where P and Q are functions of y is called a linear equation in x.
The solution can simply be written by interchanging the role of x and y.
is the solution for the linear equation (2)
Working procedure for problems
- The given equation must be first put in the form conformal to the standard form of the linear equation in x or y.
- The expression for P and Q is to be written by simple comparison.
- (^) We assume the associated solution and we only need to tackle the R.H.S. part of the solution to finally arrive at the required solution.
Example – 1
Answers:
Session – 7
Orthogonal Trajectories
Definition : If two family of curves are such that every member of one family intersects every member of the other family at right angles then they are said to be orthogonal trajectories of each other.
Working procedure for problems
Case – i : (Cartesian family)
- Given f ( x, y, c ) = 0, differentiate w.r.t x and eliminate c.
Case – ii : (Polar family)
- Given an equation in r and θ, we prefer to take logarithms first and then differentiate w.r.t θ.
Example –
Find the Orthogonal trajectories of the family of parabolas y^2 = 4^ a x.
Now differentiating (1) w.r.t x we have,
Example – 2
Example-
Show that the family of parabolas y^2 = 4 a ( x+a ) is a self orthogonal.
Consider y^2 =^^4 a^ ( x+a )^ ….(1) Differentiating w.r.t x, we have
Substituting this value of ‘a’ in (1) we have,
This is the D.E. of the given family. Now replacing y 1 by 1 / y (^) 1 (2) becomes
(3) the D.E. of the orthogonal family which is same as (2) being the D.E. of the given family. Thus the family of parabolas y^2 = 4 a ( x+a ) is self orthogonal.
Example –
Find the O.T. of the family r =a (1+sin θ) >> We have r = a (1 + sin θ)
Differentiating w.r.t θ we have,
Example – 5 Find the O.T. of the family rn^ cos^ n θ^ =^ a^ n
We have r n^ cos n θ = an
Differentiating w.r.t θ we have,
EXERCISES
Find the orthogonal trajectories of the following family of curves
LESSON – 2 Part – C Integral Calculus
Session - 1
Introduction :
We are familiar with various methods of integration, definite integrals and the associated application of finding the area under a curve.
In this chapter we first discuss reduction formulae and later discuss the method of tracing cartesian and polar curves. By knowing the shape of a given curve we disucss application of definite integrals such as area, length or perimeter, surface area of plane curves and volume of solids.
In all these applications reduction formulae plays a vital role in the evaluation of definite integrals.
Reduction Formulae
integers) to lower degree. The successive application of the recurrence relation finally end up with a function of degree 0 or 1 so that we can easily complete the integration process.
We discuss certain standard reduction formulae in the form of indefinite integrals and the evaluation of these with standard limits of integration.
We have the rule of integration by parts,
This is the required reduction formula.
Illustration
Comparing with L.H.S. of (1), we need to take n = 4 and use the established result.
We need to apply the result (1) again by taking n = 2
We cannot find I (^) 0 from (1). But basically we have
Corollary :
Equation (1) must be established first.
But cos (π / 2) = 0 = sin 0
We use this recurrence relation to find I (^) n 2 by simply replacing n by ( n 2).
Thus we have,
Note :
The result is as follows.
Session – 2
Illustration
Integrating by parts we have,
Example –
Example – 3
Example – 4
Example – 5
Exercises :
Evaluate the following integrals
Answers
Session –
Tracing of Curves
Introduction :
This topic gives an insight to the process of finding the shape of a plane curve based on its equation by examining certain features. Based on these features we can draw a rough sketch of the curve. It is highly essential to known the shape of the curve to find its area, length, surface area and volume of solids.
List of important points to be examined for tracing a cartesian curve f ( x, y ) = 0
- Symmetry : If the given equation has even powers of x only then the curve is symmetrical about the y axis and if the given equation has even powers of y only then the curve is symmetrical about the x -axis.
If f ( x, y ) = f ( y, x ) then the curve is symmetrical about the line y = x. Also if f ( x, y ) = f ( x, y ) then the curve is symmetrical about the origin.
- Special points on the curve : If f (0, 0) = 0 then the curve passes through the origin. In such a case we can find the equations of tangents at the origin by equating the groups of lowest degree terms in x and y to zero.
The points of intersection of the curve with the x -axis is got by putting y = 0 and that with the y -axis is got by putting x = 0.
- Asymptotes : Asymptote of a given curve is defined to be the tangent to the given curve at infinity. In otherwords these are lines touching the curve at infinity. Equating the coefficient of highest degree terms in x to zero we get asymptotes parallel to the x -axis and equating the coefficient of highest degree terms in y to zero we get asymptotes parallel to the y- axis.
- Region of existence : Region of existence can be determined by finding out the set of permissible (real) values x and y. The curve does not lie in the region whenever x or y is imaginary.
By examining these features we can draw a rough sketch of the curve.
Note : In the case of a parametric curve : x = x ( t ) and y = y ( t ), we need to vary the parameter t suitably to take a note of the variations in x and y so that the curve can be drawn accordingly.
List of important points to be examined for tracing a polar curve f ( r, θ) = 0
- Symmetry : f ( r, θ) = f ( r, θ) then the curve is symmetrical about the initial line
θ = 0 and θ = π.
If f ( r, θ) = f ( r, πθ) then the curve is symmetrical about the line θ = π / 2 (positive y -axis)
If f ( r, θ) = f ( r, π / 2θ) then the curve is symmetrical about the line θ = π / 4 (the line y = x )
