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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Hello dichalao

  4. anonymous
    • 5 years ago
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    Hello loki~ haha

  5. anonymous
    • 5 years ago
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    Isn't it ~1am there?

  6. anonymous
    • 5 years ago
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    yep :)

  7. anonymous
    • 5 years ago
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    i dont want to end my day as usual :(

  8. anonymous
    • 5 years ago
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    Why's that?

  9. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    I don know.. i just dont want to go to bed

  11. anonymous
    • 5 years ago
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    It should be a exponential function right

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    how you got λ2−1=0→λ=±1

  14. anonymous
    • 5 years ago
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    \[(e^{\lambda x})''-e^{\lambda x}=0 \rightarrow {\lambda}^2e^{\lambda x}-e^{\lambda x}=0 \rightarrow ({\lambda}^2-1)e^{\lambda x}=0\]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
    • 5 years ago
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    because λ=+-1 you got ce^x+c2e^(-x)?

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Last bit wasn't explained all that well.

  18. anonymous
    • 5 years ago
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    LOL it is good enough at my level

  19. anonymous
    • 5 years ago
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    Happy now? You can hit the sheets!

  20. anonymous
    • 5 years ago
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    LOL i am always happy when you are on :) flattery gets me to anywehre

  21. anonymous
    • 5 years ago
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    So do you just have to tell that guy the solution?

  22. anonymous
    • 5 years ago
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    you can do it :) i dont want to take any credits for it.. it is 10 thread up

  23. anonymous
    • 5 years ago
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    You can just screen shot it and send as an attachment.

  24. anonymous
    • 5 years ago
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    He just has to solve for the boundary conditions u(0)=0 and u(1)=1.

  25. anonymous
    • 5 years ago
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    LOL can you please do that for me :P

  26. anonymous
    • 5 years ago
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    so you will earn your 133th fan

  27. anonymous
    • 5 years ago
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    Okay, I'll go over to him. I think nikvist might be dealing with it...

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