anyone know how to solve linearly independent solutions?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

anyone know how to solve linearly independent solutions?

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

yes

- anonymous

how do I use the Wronskian method to show that f1(x) = sinh(ax) and f2(x) = cosh(ax) are linearly independent?

- anonymous

Oh, you use the definition of the Wronskian and if the determinant you get is non-zero, the solutions are linearly independent.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

The first row will be f1, f2, and the second row will have f1' and f2'. Then the determinant is f1f2'-f2f1'

- 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

does that make sense?

- anonymous

\[\sinh(ax)(\cosh(ax))'-\cosh(ax)(\sinh(ax))'=\sinh^2(ax)-\cosh^2(ax)=-1\]

- anonymous

a>0

- 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

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

- anonymous

is this the only method, or easiest method, so solving these kind of questions? L.I equations?

- anonymous

to*

- anonymous

Just one sec...distracted.

- anonymous

Are you sure the 'a' in the equation isn't a^2?

- anonymous

sorry - i just checked...off sick today

- anonymous

Haha uhh yes, its a^2

- anonymous

Whoops

- 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

Or are you asking if this is the easiest way to show linear independence?

- anonymous

If it's the latter, I would say, 'yes'.

- anonymous

yes, I meant the latter

- anonymous

And okay, thanks! I am just learning this stuff, clearly. Newb

- anonymous

No probs. Increase my fan base - become a fan!

- anonymous

Sure thing, are you a professor? Incredibly smart student?

- anonymous

Little from column A, little from column B - postgrad.

- anonymous

Cool. Well I have no idea how to do anything on here, how do I become your fan?

- anonymous

And, why does everything I type show up five hundred times?

- anonymous

There should be a blue link next to my name that says, "Become a fan".

- anonymous

Next to a blue 'thumbs up' icon.

- anonymous

Hmm... nope

- anonymous

Try refreshing the page...I hear that works sometimes.

- 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

Ah! Found it

- anonymous

awesome

- anonymous

there. have a good one!

- anonymous

you too

- anonymous

nice

Looking for something else?

Not the answer you are looking for? Search for more explanations.