anonymous 5 years ago dy/dx=cos^2(x)cos^2(2y), y=?

1. anonymous

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

2. anonymous

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

3. anonymous

Hello dichalao

4. anonymous

Hello loki~ haha

5. anonymous

Isn't it ~1am there?

6. anonymous

yep :)

7. anonymous

i dont want to end my day as usual :(

8. anonymous

Why's that?

9. anonymous

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

10. anonymous

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

11. anonymous

It should be a exponential function right

12. anonymous

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$y=c_1 e^x + c_2 e^{-x}$

13. anonymous

how you got λ2−1=0→λ=±1

14. anonymous

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

15. anonymous

because λ=+-1 you got ce^x+c2e^(-x)?

16. anonymous

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.

17. anonymous

Last bit wasn't explained all that well.

18. anonymous

LOL it is good enough at my level

19. anonymous

Happy now? You can hit the sheets!

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

LOL can you please do that for me :P

26. anonymous

so you will earn your 133th fan

27. anonymous

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