Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
theEric
Group Title
Hi! I am solving a problem, and I think I correctly arrived at \(\ln\lefte^y2x\ y\right=C\). At any rate, I think I'll be left with \(\ln\lefte^y2x\ y\right\), and I don't know how to solve for \(y\)!
 11 months ago
 11 months ago
theEric Group Title
Hi! I am solving a problem, and I think I correctly arrived at \(\ln\lefte^y2x\ y\right=C\). At any rate, I think I'll be left with \(\ln\lefte^y2x\ y\right\), and I don't know how to solve for \(y\)!
 11 months ago
 11 months ago

This Question is Closed

theEric Group TitleBest ResponseYou've already chosen the best response.0
What I have done is this:
 11 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Put each side as the exponent of \(e\). Then \(e^{\ln\lefte^y2x\ y\right}=e^C\\\implies e^y2x\ y=C\) where \(C\) is still just an arbitrary constant.
 11 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
Let me guess, you solving a ODE? Not always it is posible to explicitly express y as function of x. Sometimes you just leave a general solution in implicit form
 11 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
This is from an ODE.... I am supposed to solve it implicitly. Have I done that, then?
 11 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Assuming I got there correctly, I mean. I don't want to ask you to check my work.
 11 months ago

myko Group TitleBest ResponseYou've already chosen the best response.1
yes, you are done
 11 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Thank you! :)
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.