A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
dy/dx=cos^2(x)cos^2(2y), y=?
anonymous
 5 years ago
dy/dx=cos^2(x)cos^2(2y), y=?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is separable. You just need to rewrite it as\[\frac{dy}{\cos^2 (2y)}=\cos^2 (2x) dx \rightarrow \int\limits_{}{} \frac{dy} {\cos^2 (2y)}=\int\limits_{}{} \cos^2 (2x) dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the fact that \[\cos 2\theta =\cos^2 \theta  \sin^2 \theta = 2 \cos^2 \theta 1 \rightarrow \cos^2\theta =\frac{\cos 2 \theta +1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont want to end my day as usual :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the exact solution to this bvp u''u=0 u(0)=0 and u(1)=1 how do you solve this problem..i was trying to help the guy, but failed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don know.. i just dont want to go to bed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should be a exponential function right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol...this is a linear homogeneous second order differential equation with constant coefficients. We assume a solution of the form \[y=e^{\lambda x}\], sub it in and solve the characteristic equation\[{\lambda}^21=0 \rightarrow \lambda = \pm 1\]The solution's then\[y=c_1 e^x + c_2 e^{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how you got λ2−1=0→λ=±1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(e^{\lambda x})''e^{\lambda x}=0 \rightarrow {\lambda}^2e^{\lambda x}e^{\lambda x}=0 \rightarrow ({\lambda}^21)e^{\lambda x}=0\]In the last product, it can only be the case that\[{\lambda}^21=0\]since\[e^{\lambda x} \neq0 \]for all x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because λ=+1 you got ce^x+c2e^(x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep, the aim is to find lambdas that will allow you to make the assumption true. There's a theorem in differential equations that tells us how many solutions there will be in a diff. equation and that those solutions are unique. For second order homogeneous (what we have here) there will be two solutions, and since we've found two independent solutions (that technically needs to be checked with something called the Wronskian, but noone usually cares with these types), and since we know a set of solutions is unique, we've found the solutions to the equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Last bit wasn't explained all that well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL it is good enough at my level

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Happy now? You can hit the sheets!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL i am always happy when you are on :) flattery gets me to anywehre

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So do you just have to tell that guy the solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do it :) i dont want to take any credits for it.. it is 10 thread up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can just screen shot it and send as an attachment.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He just has to solve for the boundary conditions u(0)=0 and u(1)=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL can you please do that for me :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you will earn your 133th fan

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I'll go over to him. I think nikvist might be dealing with it...
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.