# Math Help - La' Hopital's Rule

1. ## La' Hopital's Rule

I am working on a project for my Differential Equations class and am stuck at the very last stage of the project. I have solved the problem up to the following equation:

y = 1/2 [ (1-x)^(1+b/a)/(1+b/a) - (1-x)^(1-b/a)/(1-b/a) ] + (a*b)/(a^2-b^2)

Now I need to find out what this equation becomes if a = b. If I set a equal to b the equation becomes:

y = 1/2 [ (1-x)^2/2 - 1/0] + a^2/0

Notice that there are two divisions by 0. This equation basically turns in to some term minus infinity plus infinity. It occurred to me that I can use La' Hopital's rule by rewriting the undefined portion of the equation in terms of the limit as a approaches b to find what its value becomes. However, I'm struggling to actually solve this problem.

y = 1/2 { 1/2 [ (1-x)^2 - 1 ] - ln(1-x) }

I just don't know how to get this answer. Any help is solving this will be much appreciated.

P.S. Sadly, my teacher only gave my class a week to do this project, and tomorrow's the deadline. I need to try to get this done tonight if at all possible.

2. I know what I did wrong!

I used the wrong equation... (Dope!)

I was struggling with this forever, and all I had to do was look two pages back in my notes and grab the much easier equation to use.

For those who are curious, that equation would be:

dy/dx = 1/2 [ (1-x)^(-b/a) - (1-x)^(b/a) ]

Where a = b.

Integrating this is much easier than what I was trying to do.

3. Originally Posted by ecMathGeek
I know what I did wrong!

I used the wrong equation... (Dope!)

I was struggling with this forever, and all I had to do was look two pages back in my notes and grab the much easier equation to use.

For those who are curious, that equation would be:

dy/dx = 1/2 [ (1-x)^(-b/a) - (1-x)^(b/a) ]

Where a = b.

Integrating this is much easier than what I was trying to do.
good