anonymous
  • anonymous
What am I doing wrong? DIFF EQ Initial Value Problem: \[\frac{ dy }{ dx } + \sqrt{1+t^2} e ^{-1}y = 0 ; y(0)=1\] First, I found the Integrating Factor: \[e^{\int\limits_{}^{}\sqrt{1+t^2}e ^{-1}dt}\] = \[e ^{{(\sqrt{1+t^2}+\sinh ^{-1}(t)})/2e}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
And I thought that I should have \[y(t) = ce ^{{(\sqrt{1+t^2}+\sinh ^{-1}(t)})/2e}\] but my book disagrees... any suggestions?
phi
  • phi
wolfram came up with a different answer for the integral http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bx%5E2%29%2Fe+dx
anonymous
  • anonymous
I keep getting disconnected while trying to type responses.

Looking for something else?

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

More answers

anonymous
  • anonymous
The book gives the answer \[y(t) = \exp[\int\limits_{0}^{t}\sqrt{1+s^2}e ^{-1}ds]\] my questions are, why are the bounds 0 to t. Is that because the intial value is y(0)=1? So if it had been y(3)=1 for example, the bounds would be 3 to t?
anonymous
  • anonymous
and also, why is it in terms of s, when the problem started with y's and t's?
anonymous
  • anonymous
None of these are explained in this book, so my guess is they are assumptions on something I should remember from calculus.
phi
  • phi
the s is a dummy value. I think they though the integral was to hard to solve, so they leave it in integral form
anonymous
  • anonymous
ya thats what I figure, it was a beast. What about the bounds on it though?
phi
  • phi
I would have to think about the bounds.
anonymous
  • anonymous
Dont worry about it, I have to head to class in a little while, for a different math class haha. Abstract Algebra is much more to my liking. Thanks for the assistance as always though!

Looking for something else?

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