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anonymous
 5 years ago
anyone know how to solve linearly independent solutions?
anonymous
 5 years ago
anyone know how to solve linearly independent solutions?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do I use the Wronskian method to show that f1(x) = sinh(ax) and f2(x) = cosh(ax) are linearly independent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, you use the definition of the Wronskian and if the determinant you get is nonzero, the solutions are linearly independent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first row will be f1, f2, and the second row will have f1' and f2'. Then the determinant is f1f2'f2f1'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and it also includes a differential equation of y''  a^2y = 0, I have to show that they are linearly independent for that equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sinh(ax)(\cosh(ax))'\cosh(ax)(\sinh(ax))'=\sinh^2(ax)\cosh^2(ax)=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0The above expression I wrote out (that go cut off) equals 1, which is nonzero. Your solutions are linearly independent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this the only method, or easiest method, so solving these kind of questions? L.I equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just one sec...distracted.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure the 'a' in the equation isn't a^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry  i just checked...off sick today

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha uhh yes, its a^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Re. your question about whether this is the easiest method to solve these equations: how are you exactly coming up with your solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or are you asking if this is the easiest way to show linear independence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it's the latter, I would say, 'yes'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I meant the latter

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And okay, thanks! I am just learning this stuff, clearly. Newb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No probs. Increase my fan base  become a fan!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure thing, are you a professor? Incredibly smart student?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Little from column A, little from column B  postgrad.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cool. Well I have no idea how to do anything on here, how do I become your fan?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And, why does everything I type show up five hundred times?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There should be a blue link next to my name that says, "Become a fan".

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Next to a blue 'thumbs up' icon.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try refreshing the page...I hear that works sometimes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah well, I'll find it sometime, gotta get back to studying. Thanks for the help! I will become your fan asap. :) Appreciate it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there. have a good one!
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