×

Close

- Applied Mathematics-1 - M-1
- Note
- 5 Topics
**20735 Views**- 82 Offline Downloads
- Uploaded 2 months ago

Touch here to read

Page-1

Topic:

1 Mathematics-l OBJECTIVE: The objective of the course Mathematics-l is to familiarize the prospective engineers with techniques in calculus, Gamma & Bita function, differential equation of first and second order, series solution of differential equations, Laplace transform. It aims to equip the students with standard concepts and tools at an intermediate to advanced level that will serve them well towards tackling more advanced level of mathematics and applications that they would find useful in their disciplines. Module-1 (8 hrs.) Asymptote, Curvature (Cartesian and polar), Gamma & Beta function, Partial differentiation, Maxima and Minima for function of two variables. Module-2 (8 hrs.) Differential Equation: First order differential equations, Separable Equation, Exact differential equation, Linear differential equation, Bernoulli’s equation application to Electrical circuits. Module-3 (9hrs.) Linear differential equation of second, Homogeneous equation with constant co-efficient, EulerCauchy equations, Solution by undetermined co-efficient, Solutions by variation of parameters, Modelling of electric circuits Module-4 (10 hrs.) Series solution of differential equations, Power series method, Legendre equation and Legendre polynomial. Bessels function and its properties. Module-5 (10 hrs.) Laplace transformation and its use in getting solution to differential equations, Convolution, Integral Equations. OUTCOMES On completion of this course, a student is able to: • Apply the knowledge of calculus, Gamma & Beta functions for analyzing engineering problems. • Solve first order differential equation analytically using standard method. • Demonstrate various physical models through higher order differential equation and solve such linear ordinary differential equation. • Obtain series solution of differential equation and explain application of Bessel’s function. • Apply Laplace problem to determine complete solution to ordinary differential equation.

Disclaimer: The lecture notes have been prepared by referring to many books and notes prepared by the teachers. This document does not claim any originality and cannot be used as a substitute for prescribed textbooks. The information presented here is merely a collection of materials by the author of the subject. This is just an additional tool for the teaching-learning process. The teachers, who teach in the class room, generally prepare lecture notes to give direction to the class. These notes are just a digital format of the same. These notes do not claim to be original and cannot be taken as a text book. These notes have been prepared to help the students of BPUT in their preparation for the examination. This is going to give them a broad idea about the curriculum.

3 Module -1 ASYMPTOTES DEFINITION A straight line is said to be an Asymptote of an infinite branch of a curve if as point P recedes to infinity along the branch the perpendicular distance of P the straight line tends to zero. Y Example: xy = 1 We can write either y 1 1 or x x y P when x 0 , y or y 0, x O M METHODS FOR FINDING ASYMPTOTES X Ist Method Let y = mx + c be an equation of a straight line which is not parallel to y-axis. If p = PM be the perpendicular distance of any point P(x,y) on the infinite branch of a given curve from the line y = mx + c, we have p y mx c Y 1 m2 p 0 as x lim p lim x y mx c x 0 1 m2 1 1 m2 P lim y mx c x lim y mx c 0 x lim y mx c 0 x c lim y mx x O M X

y c m x x Taking the limit as x , we get Also y = mx + c or lim x y c lim m m 0 x x x Therefore m xlim y y mx c and lim x x Working Rule y 1. Divide the equation by the highest power ‘x’ and write the equation in the form x 2. Taking the limit as x , it implies m xlim y x 3. Write the equation in the form (y – mx) and taking the limit x y mx c 4. Put lim x 5. Put the values of m and c in the equation y = mx + c to get the required asymptotes. Example: x 3 y 3 3axy 0 Divide term x3 i.e. highest power, we get 3 y y 1 1 3a 0 x x x Taking x , we get 2 1 m3 0 1 m 1 m m 0 m 1 Other values are imaginary, Next, we have to calculate c lim y mx , c lim y x x x 3 3 So, we write x y 3axy 0 xy 3axy x xy y 2 lim x y lim 3axy x xy y 2 x x 2 2 y 3a 3a 1 x c lim a 2 2 x 1 y y 1 1 1 x x

## Leave your Comments

## Akhilesh tiwary

5 days ago00## tarunya L

21 days ago10## NITISH SINGH

1 month ago14