# Linear initial-value question

1. Sep 8, 2009

### string_656

Im doing questions out of a book, and im stuck on one question, and it dosnt give any answers

dy/dx*(x) = 2y/x + X^2.......when y(1) = 3 ... = ..... dy/dx - 2y/x^2 = x
i have trouble intergrating...

I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

e^(2/x)*y = (intergral of) xe^(2/x)

im just having trouble with this ^^

can ya help?
thanks

2. Sep 9, 2009

### LCKurtz

Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere.

3. Sep 9, 2009

### HallsofIvy

Staff Emeritus
On the contrary, that has a simple polynomial solution.

Your difficulty is in your very first step: your equation is dy/dx= 2y/x+ x2 and then you rewrote it as dy/dx- 2y/x2= x!

It should be dy/dx- 2y/x= x2. Now the integrating factor is
$$e^{\int \frac{-2}{x}dx}= x^{-2}$$.

4. Sep 9, 2009

### LCKurtz

But aren't you missing the x in the first term of the original DE: dy/dx*(x)?
I would have written it as x*dy/dx but anyway you must divide through by that before computing the integrating factor.

5. Sep 9, 2009

### string_656

im only having trouble intergrating xe^(2/x).. if i can do that i can solve the problem..

6. Sep 10, 2009

### LCKurtz

In my previous reply I said:

"Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere."

There *is* no elementary anti-derivative for that function. None of your usual techniques can ever work.

7. Sep 10, 2009

### string_656

oh ok.. thanks. but how could i do it... is there a way of doing it?

8. Sep 11, 2009

### LCKurtz

There is no elementary way. If you are familiar with the exponential integral:

http://en.wikipedia.org/wiki/Exponential_integral

there is a way to express the answer in terms of that. Like I said before, if this is a typical homework problem I suspect there is a misprint. As another poster has pointed out, if the x factor in the leading term was missing, it would be straightforward. I can give you the expression Maple gives for an answer if you want it.

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