anonymous
  • anonymous
anyone know how to solve linearly independent solutions?
Mathematics
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anonymous
  • anonymous
anyone know how to solve linearly independent solutions?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
yes
anonymous
  • anonymous
how do I use the Wronskian method to show that f1(x) = sinh(ax) and f2(x) = cosh(ax) are linearly independent?
anonymous
  • anonymous
Oh, you use the definition of the Wronskian and if the determinant you get is non-zero, the solutions are linearly independent.

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anonymous
  • anonymous
The first row will be f1, f2, and the second row will have f1' and f2'. Then the determinant is f1f2'-f2f1'
anonymous
  • anonymous
and it also includes a differential equation of y'' - a^2y = 0, I have to show that they are linearly independent for that equation
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
\[\sinh(ax)(\cosh(ax))'-\cosh(ax)(\sinh(ax))'=\sinh^2(ax)-\cosh^2(ax)=-1\]
anonymous
  • anonymous
a>0
anonymous
  • anonymous
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.
anonymous
  • anonymous
The above expression I wrote out (that go cut off) equals -1, which is non-zero. Your solutions are linearly independent.
anonymous
  • anonymous
is this the only method, or easiest method, so solving these kind of questions? L.I equations?
anonymous
  • anonymous
to*
anonymous
  • anonymous
Just one sec...distracted.
anonymous
  • anonymous
Are you sure the 'a' in the equation isn't a^2?
anonymous
  • anonymous
sorry - i just checked...off sick today
anonymous
  • anonymous
Haha uhh yes, its a^2
anonymous
  • anonymous
Whoops
anonymous
  • anonymous
Re. your question about whether this is the easiest method to solve these equations: how are you exactly coming up with your solutions?
anonymous
  • anonymous
Or are you asking if this is the easiest way to show linear independence?
anonymous
  • anonymous
If it's the latter, I would say, 'yes'.
anonymous
  • anonymous
yes, I meant the latter
anonymous
  • anonymous
And okay, thanks! I am just learning this stuff, clearly. Newb
anonymous
  • anonymous
No probs. Increase my fan base - become a fan!
anonymous
  • anonymous
Sure thing, are you a professor? Incredibly smart student?
anonymous
  • anonymous
Little from column A, little from column B - postgrad.
anonymous
  • anonymous
Cool. Well I have no idea how to do anything on here, how do I become your fan?
anonymous
  • anonymous
And, why does everything I type show up five hundred times?
anonymous
  • anonymous
There should be a blue link next to my name that says, "Become a fan".
anonymous
  • anonymous
Next to a blue 'thumbs up' icon.
anonymous
  • anonymous
Hmm... nope
anonymous
  • anonymous
Try refreshing the page...I hear that works sometimes.
anonymous
  • anonymous
Ah well, I'll find it sometime, gotta get back to studying. Thanks for the help! I will become your fan asap. :) Appreciate it.
anonymous
  • anonymous
Ah! Found it
anonymous
  • anonymous
awesome
anonymous
  • anonymous
there. have a good one!
anonymous
  • anonymous
you too
anonymous
  • anonymous
nice

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