# More linear eqn trouble

1. Jun 20, 2004

### Math Is Hard

Staff Emeritus
My teacher gave us this problem in class and then when she went to show us how to solve, she got stuck too!

$$1+xy = xy'$$

after some rearranging and dividing we have

$$y' - y = 1/x$$

we used an integrating factor of $$e^ {-x}$$

and got the result of $$(e^ {-x}y)' = e^ {-x}/x$$

integrating gives $$(e^ {-x}y) = {\int e^ {-x}/x \dx}$$

and that's about where we got stuck. Integration by parts just took us in a loop.
Any help is appreciated. Thanks!

2. Jun 20, 2004

### Gokul43201

Staff Emeritus
I think you/she got stuck there because the integral of $${e^ {-x}/x \dx}$$ has no analytic form - it diverges at x=0

Last edited: Jun 20, 2004
3. Jun 20, 2004

### Math Is Hard

Staff Emeritus
Thanks, Gokul.

so as far as the solution for y , do you just throw up your hands and say "OK, divergent, can't be solved since the integral won't converge" or can you express y as something like this and call that the solution?

$$y = {e^ {x}\int e^ {-x}/x \dx}$$

4. Jun 20, 2004

### Tom Mattson

Staff Emeritus
You are correct about both of those statements, but I think I should clarify by saying that one has nothing to do with the other. The integral of exp(-x2) has no analytic form, and it doesn't diverge anywhere. Conversely, the integral of tan(x) does have an analytic form, and it diverges at (2n+1)&pi;/2, for any integer n.

Math Is Hard,

Yes, you can leave an integral as part of a solution.

5. Jun 20, 2004

### Math Is Hard

Staff Emeritus
Thanks, Tom.