A community for students.
Here's the question you clicked on:
 0 viewing
gerryliyana
 2 years ago
solve y dx + (3xy  1) dy = 0
gerryliyana
 2 years ago
solve y dx + (3xy  1) dy = 0

This Question is Closed

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1\[y dx + (3xy  1 ) dy = 0\] \[y dx + 3xy dy  dy = 0\]\[y \frac{ dx }{ dy }+3xy  1 = 0\] \[\frac{ dx }{ dy }+ 3x = \frac{ 1 }{ y }\] I think, in order to satisfy the equation \(\frac{ dx }{ dy } + p(y)x = g(y) x\) \(p(y) = 3\) and \(g(y) = \frac{ 1 }{ y }\) \[\mu(y) = \exp(\int\limits\limits_{0}^{y} p(y) dy) = \exp (\int\limits\limits_{0}^{y} 3 dy) = e^{3y}\] \[x = \frac{ 1 }{ \mu(y) }\int\limits_{0}^{y} \mu(y) g(y) dy\] \[x = \frac{ 1 }{ e^{3y} } \int\limits_{0}^{y} e^{3y} \frac{ 1 }{ y } dy\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1I got \[xe^{3y}  \int\limits \frac{ e^{3y} }{ y } = C\] for the solution.

Kira_Yamato
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not too sure, sorry.... But I think it should be ok...

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1Ok, np @Kira_Yamato i'm not sure for: \[\int\limits \frac{ e^{3y} }{ y } dy\] what results did you get?

Kira_Yamato
 2 years ago
Best ResponseYou've already chosen the best response.0This is what MatLab gave me

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1can i see ur script on matlab??

Kira_Yamato
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry I mean Wolfram Alpha

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1ok.., thank u Kira :)
Ask your own question
Sign UpFind more explanations on OpenStudy
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.