Non-homogenous secx ODE's and Euler eq's

  • Thread starter Thread starter hivesaeed4
  • Start date Start date
  • Tags Tags
    Euler
hivesaeed4
Messages
217
Reaction score
0
Suppose we have Asec(x) on the right hand side in a non-homogenous ODE and in a Euler equation. How do we solve it? ( I know how to solve for cos and sin on the right hand side but not for any other trig function).
 
Physics news on Phys.org
Help?
 
For non-homogeneous ordinary differential equations, i was taught that you always had to use the method of annihilators if the right hand side was either cos(x), sin(x), exp(x), a polynomial function or the product and sum of any of these functions. For functions like sec(x)=1/cos(x) the annihilator method won't work, and therefore you will need to use the variation of parameters method to solve your differential equation.

It's how i was taught, so i don't know if there is another method out there that could be used.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top