dy/dx=cos^2(x)cos^2(2y), y=?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

dy/dx=cos^2(x)cos^2(2y), y=?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

This is separable. You just need to rewrite it as\[\frac{dy}{\cos^2 (2y)}=\cos^2 (2x) dx \rightarrow \int\limits_{}{} \frac{dy} {\cos^2 (2y)}=\int\limits_{}{} \cos^2 (2x) dx\]
Use the fact that \[\cos 2\theta =\cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta -1 \rightarrow \cos^2\theta =\frac{\cos 2 \theta +1}{2}\]
Hello dichalao

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hello loki~ haha
Isn't it ~1am there?
yep :)
i dont want to end my day as usual :(
Why's that?
find 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
I don know.. i just dont want to go to bed
It should be a exponential function right
Lol...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\[{\lambda}^2-1=0 \rightarrow \lambda = \pm 1\]The solution's then\[y=c_1 e^x + c_2 e^{-x}\]
how you got λ2−1=0→λ=±1
\[(e^{\lambda x})''-e^{\lambda x}=0 \rightarrow {\lambda}^2e^{\lambda x}-e^{\lambda x}=0 \rightarrow ({\lambda}^2-1)e^{\lambda x}=0\]In the last product, it can only be the case that\[{\lambda}^2-1=0\]since\[e^{\lambda x} \neq0 \]for all x.
because λ=+-1 you got ce^x+c2e^(-x)?
Yep, 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 no-one usually cares with these types), and since we know a set of solutions is unique, we've found the solutions to the equation.
Last bit wasn't explained all that well.
LOL it is good enough at my level
Happy now? You can hit the sheets!
LOL i am always happy when you are on :) flattery gets me to anywehre
So do you just have to tell that guy the solution?
you can do it :) i dont want to take any credits for it.. it is 10 thread up
You can just screen shot it and send as an attachment.
He just has to solve for the boundary conditions u(0)=0 and u(1)=1.
LOL can you please do that for me :P
so you will earn your 133th fan
Okay, I'll go over to him. I think nikvist might be dealing with it...

Not the answer you are looking for?

Search for more explanations.

Ask your own question