A community for students.
Here's the question you clicked on:
 0 viewing
gazc99
 2 years ago
Diff eqn dy/dx= (ycos(x))/(1+2y^2) int condt y(0)=1
gazc99
 2 years ago
Diff eqn dy/dx= (ycos(x))/(1+2y^2) int condt y(0)=1

This Question is Open

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1this is easily separable. bring it in the form f(x)dx = f(y)dy then integrate both sides.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1can you ? or need help with that ?

Mimi_x3
 2 years ago
Best ResponseYou've already chosen the best response.2because im bored...\[\frac{dy}{dx} = \frac{ycos(x)}{1+2y^{2}} \] \[=>\frac{dy}{dx} = \frac{y}{1+2y^{2}} *\cos(x) => dy = \frac{y}{1+2y^{2}} *\cos(x)dx => \frac{dy(1+2y^{2})}{y} = cos(x) \] \[=> \int\limits\frac{1+2y^{2}}{y} dy = \int\limits \cos(x) \] \[\int\limits\frac{1}{y} dy + \frac{2y^{2}}{y} dy = \ln(y) + \frac{y^{2}}{2} \] \[=> \ln(y)+\frac{y^{2}}{2} = sinx +c\] very easy DE..

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0hmm, it's \[ \ln y+y^2=\sin x + C\] now put \(x=0\) and \(y=1\) to find \(C\)

Mimi_x3
 2 years ago
Best ResponseYou've already chosen the best response.2wait there's a typo..\[ \int\limits\frac{2y^{2}}{y} dy => \int\limits2y dy => y^2\]
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.