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: $$e^{\int\limits_{}^{}\sqrt{1+t^2}e ^{-1}dt}$$ = $$e ^{{(\sqrt{1+t^2}+\sinh ^{-1}(t)})/2e}$$

Question

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: $$e^{\int\limits_{}^{}\sqrt{1+t^2}e ^{-1}dt}$$ = $$e ^{{(\sqrt{1+t^2}+\sinh ^{-1}(t)})/2e}$$

anonymous · Answer

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 · Answer

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 · Answer

I keep getting disconnected while trying to type responses.

anonymous · Answer

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 · Answer

and also, why is it in terms of s, when the problem started with y's and t's?

anonymous · Answer

None of these are explained in this book, so my guess is they are assumptions on something I should remember from calculus.

phi · Answer

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 · Answer

ya thats what I figure, it was a beast. What about the bounds on it though?

phi · Answer

I would have to think about the bounds.

anonymous · Answer

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!