How can I solve this linear equation with an integrating factor?

[Math Is Hard]

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!

 

[Gokul43201]

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!

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

 

[Math Is Hard]

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}

 

[quantumdude]

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

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)π/2, for any integer n.

Math Is Hard,

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

 

[Math Is Hard]

Thanks, Tom.