Question

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

Answer 1

This is separable. You just need to rewrite it as\[\frac{dy}{\cos^2 (2y)}=\cos^2 (2x) dx \rightarrow \int\limits_{}{} \frac{dy} {\cos^2 (2y)}=\int\limits_{}{} \cos^2 (2x) dx\]

Answer 2

Use the fact that \[\cos 2\theta =\cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta -1 \rightarrow \cos^2\theta =\frac{\cos 2 \theta +1}{2}\]

Answer 3

Hello dichalao

Answer 4

Hello loki~ haha

Answer 5

Isn't it ~1am there?

Answer 6

yep :)

Answer 7

i dont want to end my day as usual :(

Answer 8

Why's that?

Answer 9

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

Answer 10

I don know.. i just dont want to go to bed

Answer 11

It should be a exponential function right

Answer 12

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\[y=c_1 e^x + c_2 e^{-x}\]

Answer 13

how you got λ2−1=0→λ=±1

Answer 14

\[(e^{\lambda x})''-e^{\lambda x}=0 \rightarrow {\lambda}^2e^{\lambda x}-e^{\lambda x}=0 \rightarrow ({\lambda}^2-1)e^{\lambda x}=0\]In the last product, it can only be the case that\[{\lambda}^2-1=0\]since\[e^{\lambda x} \neq0 \]for all x.

Answer 15

because λ=+-1 you got ce^x+c2e^(-x)?

Answer 16

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.

Answer 17

Last bit wasn't explained all that well.

Answer 18

LOL it is good enough at my level

Answer 19

Happy now? You can hit the sheets!

Answer 20

LOL i am always happy when you are on :) flattery gets me to anywehre

Answer 21

So do you just have to tell that guy the solution?

Answer 22

you can do it :) i dont want to take any credits for it.. it is 10 thread up

Answer 23

You can just screen shot it and send as an attachment.

Answer 24

He just has to solve for the boundary conditions u(0)=0 and u(1)=1.

Answer 25

LOL can you please do that for me :P

Answer 26

so you will earn your 133th fan

Answer 27

Okay, I'll go over to him. I think nikvist might be dealing with it...