Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
henpen
Group Title
I'm not sure if this is even solvable, but I'd like to know what sort of things you would need to solve this: \[y''+k(y')^2y=0 \]
 one year ago
 one year ago
henpen Group Title
I'm not sure if this is even solvable, but I'd like to know what sort of things you would need to solve this: \[y''+k(y')^2y=0 \]
 one year ago
 one year ago

This Question is Closed

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
What are you looking for?
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
I'm guessing y, but you didn't specify.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351369287726:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351369416936:dw
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Nice. I was only able to see y=c as the trivial solution.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
there are two answers that one of them is y=c1
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
I knew there was another one, but I was stuck after y=c. :">
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Thank you for that, sorry for being ambiguous in the question.
 one year 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.