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anonymous

  • 5 years ago

anyone know how to solve linearly independent solutions?

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  1. anonymous
    • 5 years ago
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    yes

  2. anonymous
    • 5 years ago
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    how do I use the Wronskian method to show that f1(x) = sinh(ax) and f2(x) = cosh(ax) are linearly independent?

  3. anonymous
    • 5 years ago
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    Oh, you use the definition of the Wronskian and if the determinant you get is non-zero, the solutions are linearly independent.

  4. anonymous
    • 5 years ago
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    The first row will be f1, f2, and the second row will have f1' and f2'. Then the determinant is f1f2'-f2f1'

  5. anonymous
    • 5 years ago
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    and it also includes a differential equation of y'' - a^2y = 0, I have to show that they are linearly independent for that equation

  6. anonymous
    • 5 years ago
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    does that make sense?

  7. anonymous
    • 5 years ago
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    \[\sinh(ax)(\cosh(ax))'-\cosh(ax)(\sinh(ax))'=\sinh^2(ax)-\cosh^2(ax)=-1\]

  8. anonymous
    • 5 years ago
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    a>0

  9. anonymous
    • 5 years ago
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    Yeah, you can use the Wronskian to show that two solutions will be linearly independent. You would have obtained those solutions from the differential equation. Then you plug them into det(W) to see if they're independent.

  10. anonymous
    • 5 years ago
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    The above expression I wrote out (that go cut off) equals -1, which is non-zero. Your solutions are linearly independent.

  11. anonymous
    • 5 years ago
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    is this the only method, or easiest method, so solving these kind of questions? L.I equations?

  12. anonymous
    • 5 years ago
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    to*

  13. anonymous
    • 5 years ago
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    Just one sec...distracted.

  14. anonymous
    • 5 years ago
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    Are you sure the 'a' in the equation isn't a^2?

  15. anonymous
    • 5 years ago
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    sorry - i just checked...off sick today

  16. anonymous
    • 5 years ago
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    Haha uhh yes, its a^2

  17. anonymous
    • 5 years ago
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    Whoops

  18. anonymous
    • 5 years ago
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    Re. your question about whether this is the easiest method to solve these equations: how are you exactly coming up with your solutions?

  19. anonymous
    • 5 years ago
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    Or are you asking if this is the easiest way to show linear independence?

  20. anonymous
    • 5 years ago
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    If it's the latter, I would say, 'yes'.

  21. anonymous
    • 5 years ago
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    yes, I meant the latter

  22. anonymous
    • 5 years ago
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    And okay, thanks! I am just learning this stuff, clearly. Newb

  23. anonymous
    • 5 years ago
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    No probs. Increase my fan base - become a fan!

  24. anonymous
    • 5 years ago
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    Sure thing, are you a professor? Incredibly smart student?

  25. anonymous
    • 5 years ago
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    Little from column A, little from column B - postgrad.

  26. anonymous
    • 5 years ago
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    Cool. Well I have no idea how to do anything on here, how do I become your fan?

  27. anonymous
    • 5 years ago
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    And, why does everything I type show up five hundred times?

  28. anonymous
    • 5 years ago
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    There should be a blue link next to my name that says, "Become a fan".

  29. anonymous
    • 5 years ago
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    Next to a blue 'thumbs up' icon.

  30. anonymous
    • 5 years ago
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    Hmm... nope

  31. anonymous
    • 5 years ago
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    Try refreshing the page...I hear that works sometimes.

  32. anonymous
    • 5 years ago
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    Ah well, I'll find it sometime, gotta get back to studying. Thanks for the help! I will become your fan asap. :) Appreciate it.

  33. anonymous
    • 5 years ago
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    Ah! Found it

  34. anonymous
    • 5 years ago
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    awesome

  35. anonymous
    • 5 years ago
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    there. have a good one!

  36. anonymous
    • 5 years ago
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    you too

  37. anonymous
    • 5 years ago
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    nice

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