Lecture

Linear Equation Solution Example

Description

This lecture demonstrates the resolution of a linear equation with a second member using previously seen formulas. Starting with known solutions without a second member, U1 and U2, the general solution is reconstructed by calculating the two gamma functions derived from the Wronskian W. The process involves calculating the Wronskian, then the gamma functions through integrals, and finally combining everything to form the general solution with the second member. The final expression includes the product of constants c1 and c2, initial solutions e³t and 2t³ x e³ –t, and an additional simplified function –t³, representing the general structure of the solution.

About this result
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website. Please make sure to verify the information with EPFL's official sources.

Graph Chatbot

Chat with Graph Search

Ask any question about EPFL courses, lectures, exercises, research, news, etc. or try the example questions below.

DISCLAIMER: The Graph Chatbot is not programmed to provide explicit or categorical answers to your questions. Rather, it transforms your questions into API requests that are distributed across the various IT services officially administered by EPFL. Its purpose is solely to collect and recommend relevant references to content that you can explore to help you answer your questions.