A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
prove y=(ln(x) + c)/x is a soln. of: x^2y' + xy = 1
anonymous
 5 years ago
prove y=(ln(x) + c)/x is a soln. of: x^2y' + xy = 1

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you post again your question? please be more clear.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=(\ln(x) + c)/x\] solution of: \[x^2y' + xy = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0prove it; \[y=(\ln(x)+c)/x\] \[xy=\ln(x)+c\] implicit derivative \[xy'+y=1/x\] multiply for x to both sides \[x^2y′+xy=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you became my fan!

apples
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2y' + xy = 1\]\[x^2 \cdot \frac{c + \log{x}  1}{x^2} + x\frac{\log{x} + c}{x} = 1\]\[\log{x} + c  (c+\log{x}1) = 1\]\[1=1\] Thus \[\frac{\log{x}+c}{x}\] is a general solution of that differential equation.
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.