# Thread: Tricky Integral (from a differential equation)

1. ## 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 with respect to y

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

Any ideas?

2. ## Re: Tricky Integral

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 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?

3. ## 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.

4. ## Re: Tricky Integral

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.

5. ## Re: Tricky Integral

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

The question was:

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!

6. ## Re: Tricky Integral

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

The question was:

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

7. ## Re: Tricky Integral

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!

8. ## Re: Tricky Integral

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

9. ## 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!

10. ## Re: Tricky Integral

The equation is linear in x. If you use the substitution

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

the DE becomes

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

Putting in standard form yields

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

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

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

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.

11. ## 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.

12. ## Re: Tricky Integral (from a differential equation)

I found out today that the integral was supposed to be $\displaystyle${sin(\sqrt{y}) which is pretty straightforward to integrate!

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