# Tricky Integral (from a differential equation)

• June 15th 2011, 01:46 PM
dtwazere
Tricky Integral (from a differential equation)
Hi,

I came across this integral in my exam the other day. I have spoken to other people and nobody was able to compute it.

The integral http://img171.imageshack.us/img171/2...0110615164.gif with respect to y

I tried by parts, and by substitution, maple and wolfram and I can't seem to do it!

Any ideas?
• June 15th 2011, 01:58 PM
mr fantastic
Re: Tricky Integral
Quote:

Originally Posted by dtwazere
Hi,

I came across this integral in my exam the other day. I have spoken to other people and nobody was able to compute it.

The integral http://img171.imageshack.us/img171/2...0110615164.gif with respect to y

I tried by parts, and by substitution, maple and wolfram and I can't seem to do it!

Any ideas?

Where has the integral come from? Was it a definite integral? Were you expected to calculate it by hand? What syllabus was the Exam examining?
• June 15th 2011, 02:01 PM
dtwazere
Re: Tricky Integral
It was from an exact equation. The exam was called 'Mathematical Methods' and involved differential equations.

It was the g'(y) at the end and we had to integrate it as the last thing to do.
• June 15th 2011, 02:04 PM
mr fantastic
Re: Tricky Integral
Quote:

Originally Posted by dtwazere
It was from an exact equation. The exam was called 'Mathematical Methods' and involved differential equations.

It was the g'(y) at the end and we had to integrate it as the last thing to do.

Perhaps you had to use an infinite series. The fact is, without seeing the whole question and putting it into context it's impossible to answer. As posted, the integral is non-elementary.
• June 15th 2011, 03:06 PM
dtwazere
Re: Tricky Integral
Hmm I'm on an undergrad maths course and infinite series were not in this module.

The question was:
http://img28.imageshack.us/img28/549...0110615180.gif

I calculated the integrating factor to be y^(-1/2)

We're all stumped! We are adamant it couldn't be solved knowing what we were taught, but we want to be sure before we raise the point.

I hope this isn't off topic as it's actually the differential equation now!
• June 15th 2011, 03:34 PM
topsquark
Re: Tricky Integral
Quote:

Originally Posted by dtwazere
Hmm I'm on an undergrad maths course and infinite series were not in this module.

The question was:
http://img28.imageshack.us/img28/549...0110615180.gif

I calculated the integrating factor to be y^(-1/2)

We're all stumped! We are adamant it couldn't be solved knowing what we were taught, but we want to be sure before we raise the point.

I hope this isn't off topic as it's actually the differential equation now!

I'm not making heads nor tails out of it. Do you have boundary conditions?

-Dan
• June 15th 2011, 03:37 PM
dtwazere
Re: Tricky Integral
Quote:

Originally Posted by topsquark
I'm not making heads nor tails out of it. Do you have boundary conditions?

-Dan

We had y(0) = Pi^2 / 9 if that helps!
• June 15th 2011, 03:56 PM
topsquark
Re: Tricky Integral
Quote:

Originally Posted by dtwazere
We had y(0) = Pi^2 / 9 if that helps!

Well I was hoping. Still, you can do this with a series approximation (not quite sure how you are going to work in the boundary condition without some advanced knowledge of series). You've surely heard of the MacLaurin series of a function. Find the MacLaurin series of your integrand then integrate term by term. The solution of your differential equation will then be in the form of x = f(y), which I doubt you will find satisfactory. Without using a power series solution method this is as good as you are likely to get.

-Dan
• June 15th 2011, 04:03 PM
dtwazere
Re: Tricky Integral
I've heard of the MacLaurin but thats about it! :P

It was a surprise in the middle of the exam, and the lecturer told us that it wasn't a calculus exam so we were not to expect to prove any integrals!

Thanks for your help, it's been bugging me for days!
• June 15th 2011, 05:08 PM
Ackbeet
Re: Tricky Integral
The equation is linear in x. If you use the substitution

$u=y^{2},\quad u'=2yy',$

the DE becomes

$(x-\sin(u))\,u'+4u=0.$

Putting in standard form yields

$\frac{dx}{du}+\frac{x}{4u}=\frac{\sin(u)}{4u}.$

The integrating factor is equivalent to what you found. However, by the uniqueness theorem, the answer you get from the integrating factor is unique on either of the intervals

$u\in(-\infty,0),\;\text{or}\; u\in(0,\infty).$

Given the initial condition, you would pick the latter. I think the best you can do with a non-series approach is to reduce to quadratures.
• June 16th 2011, 03:37 AM
HallsofIvy
Re: Tricky Integral (from a differential equation)
While attempting the first homework set in the first course I ever took in differential equations, I was able to write the solution as an integral but simply was not able to do the integral. Finally, since this was a problem that had the answer in the back of the book, I looked there to discover that the solution was given in the form of an integral! Perhaps that is what is expected here.
• June 20th 2011, 12:07 PM
dtwazere
Re: Tricky Integral (from a differential equation)
I found out today that the integral was supposed to be ${sin(\sqrt{y})$ which is pretty straightforward to integrate!

Thanks to everyone who tried that integral, I learnt something new (Rofl)