Calculus and Analysis Assignment - Course Name, Semester 1, 2024

This document provides a comprehensive solution to a calculus and analysis assignment. It covers several key concepts, including definite integrals, integration by parts, and differential equations. The solution demonstrates the application of these concepts through detailed step-by-step calculations and explanations. The assignment involves solving for the inverse of a matrix, analyzing systems of linear equations, and finding both general and particular solutions to differential equations. The solution also covers the constant solution of a differential equation and demonstrates techniques for solving separable differential equations. Overall, the assignment provides a thorough exploration of fundamental calculus and analysis principles, offering a valuable resource for students seeking to understand and master these topics. This assignment is contributed by a student to be published on the website Desklib. Desklib is a platform which provides all the necessary AI based study tools for students.

Solution Task 1:
(a)
Consider
Since , let’s consider a particular case , then
Simplify further,
Hence
(b)
Consider the limit
Since
So,
Hence
Solution Task 2:
(a)
Consider the definite integral where and are positive
constants.
Now,
Simplify further,
Simplify further,

Hence
(b)
To show
Last result from part (a)
This completes the proof.
Solution Task 3:
Consider the system of equation,
(a)
The augmented matrix is,
Now let’s perform elementary row operations.
Perform
(a)
Since system has four equations and three unknown that is system has either infinitely
many solutions or no solution.
Now,
If that is for system has infinitely many solutions and if that is
if then the system is inconsistent
(b)
If then system has infinitely many solutions that is

Suppose then
Where
Solution Task 4:
Given the matrices,
(a)
That is
And
That is
(b)
Since hence inverse of exist.
Now the augmented matrix with identity is,
Now let’s perform elementary operations to the augmented matrix.

Hence, the inverse of matrix is
(c)
Since
That is,
Since this implies that
That is and must be same.
Solution Task 5:
(a)
To prove
Consider the integral
Use integral by part method,
Simplify further,
Hence,
This completes the proof.
(b)
Consider the differential equation
Now,
Implies that

This is a linear ordinary differential equation of the form whose
solution is given by where is an integrating
factor.
Here so integrating factor
Then the general solution is,
This gives
(c)
Since use this in above equation, we get
Then the required particular solution is
Solution Task 6:
Consider the differential equation this gives,
(a)
Note that a constant solution of differential equation occurs when derivative will be zero
That is
Since this implies that
Hence differential equation has a constant solution at
(b)
Since this implies that
This is of variable separable form so integrate and obtain