anonymous
  • anonymous
dy/dx=cos^2(x)cos^2(2y), y=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This is separable. You just need to rewrite it as\[\frac{dy}{\cos^2 (2y)}=\cos^2 (2x) dx \rightarrow \int\limits_{}{} \frac{dy} {\cos^2 (2y)}=\int\limits_{}{} \cos^2 (2x) dx\]
anonymous
  • anonymous
Use the fact that \[\cos 2\theta =\cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta -1 \rightarrow \cos^2\theta =\frac{\cos 2 \theta +1}{2}\]
anonymous
  • anonymous
Hello dichalao

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anonymous
  • anonymous
Hello loki~ haha
anonymous
  • anonymous
Isn't it ~1am there?
anonymous
  • anonymous
yep :)
anonymous
  • anonymous
i dont want to end my day as usual :(
anonymous
  • anonymous
Why's that?
anonymous
  • 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
anonymous
  • anonymous
I don know.. i just dont want to go to bed
anonymous
  • anonymous
It should be a exponential function right
anonymous
  • 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\[y=c_1 e^x + c_2 e^{-x}\]
anonymous
  • anonymous
how you got λ2−1=0→λ=±1
anonymous
  • 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.
anonymous
  • anonymous
because λ=+-1 you got ce^x+c2e^(-x)?
anonymous
  • 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.
anonymous
  • anonymous
Last bit wasn't explained all that well.
anonymous
  • anonymous
LOL it is good enough at my level
anonymous
  • anonymous
Happy now? You can hit the sheets!
anonymous
  • anonymous
LOL i am always happy when you are on :) flattery gets me to anywehre
anonymous
  • anonymous
So do you just have to tell that guy the solution?
anonymous
  • anonymous
you can do it :) i dont want to take any credits for it.. it is 10 thread up
anonymous
  • anonymous
You can just screen shot it and send as an attachment.
anonymous
  • anonymous
He just has to solve for the boundary conditions u(0)=0 and u(1)=1.
anonymous
  • anonymous
LOL can you please do that for me :P
anonymous
  • anonymous
so you will earn your 133th fan
anonymous
  • anonymous
Okay, I'll go over to him. I think nikvist might be dealing with it...

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