Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Find the root of the given equation in terms of ##u##
Reply to thread
Message
[QUOTE="chwala, post: 6870224, member: 287397"] aaargh i get you [USER=572553]@fresh_42[/USER] ...should have double checked... i will amend my response in post ##5## as follows ...just in following with your working, ##x=\dfrac{-1±\sqrt{1+ \sinh^2 2u}}{\sinh 2u}=\dfrac{-1±\sqrt{\cosh^2 2u}}{\sinh 2u}=\dfrac{-1±\cosh 2u}{\sinh 2u}=\dfrac{-1±(1+2\sinh^{2} u)}{2\sinh u \cosh u}## Now, we have two possibilities; ##x_1=\dfrac{-1+1+2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{\sinh u}{\cosh u}## i can also see different variations on this by you and also by [USER=147785]@Mark44[/USER] to be specific; ##\dfrac{\sinh u}{\cosh u}=\tanh u##. Also, ##x_2=\dfrac{-1-1-2\sinh^{2} u}{2\sinh u \cosh u}=\dfrac{-2-2\sinh^{2} u}{2\sinh u \cosh u}=-\dfrac{2(1+\sinh^{2} u)}{2\sinh u \cosh u}=-\dfrac{1+\sinh^{2} u}{\sinh u \cosh u}## ##=-\dfrac{\cosh^{2} u}{\sinh u \cosh u}=-\dfrac{\cosh u}{\sinh u }## Thanks [USER=572553]@fresh_42[/USER] , i actually did not go through earlier posts in detail because of other work related duties...regards, [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Find the root of the given equation in terms of ##u##
Back
Top