# Homework Help: Find y' if x[SUP]5y[/SUP]=y[SUP]2x[/SUP]

1. Mar 14, 2013

### catherinenanc

1. The problem statement, all variables and given/known data

Find y' if x5y=y2x.

2. Relevant equations

d(ln y)=dy/y.

3. The attempt at a solution

I know this involves the natural log, and I suspect I am supposed to take the ln of both sides in order to pop the exponents down in front, and then use d(ln y)=dy/y. But I can't really complete the thought. It's been a while!

2. Mar 14, 2013

### Staff: Mentor

What do you get when you take the log (natural) of each side?
IOW, what is ln(x5y)? What is ln(y2x)?

3. Mar 14, 2013

### catherinenanc

I'm pretty sure you get (5y)(ln x)=(2x)(ln y), right?

4. Mar 14, 2013

### catherinenanc

I know you can't isolate y, but I could atleast gather the y's on the left and the x's on the right, which gives me (5y)/(ln y)=(2x)/(ln x).

5. Mar 14, 2013

### catherinenanc

Maybe instead I want (ln y)/(5y)=(ln x)/(2x), because then I have 5f(y)dy=2f(x)dx where f is the ln function.

6. Mar 14, 2013

### Staff: Mentor

Right.
You don't need to isolate y. Just differentiate both sides of the first equation above implicitly.

7. Mar 14, 2013

### catherinenanc

(5y)(dx/x) + (ln x)(5dy)=(2x)(dy/y) + (ln y)(2dx)
(ln x)(5dy) - (2x)(dy/y)=(ln y)(2dx) - (5y)(dx/x)
[5(ln x) - 2x/y]dy=[2(ln y) - 5y/x]dx
dy/dx=[2(ln y) - 5y/x]/[5(ln x) - 2x/y]

8. Mar 14, 2013

### catherinenanc

but i still get dy/dx in terms of y. Aren't I suppose to find it just in terms of x?

9. Mar 14, 2013

### Staff: Mentor

This looks fine.

All it asks is that you find y'. It doesn't say anything about it needing to be as a function of x alone.

10. Mar 14, 2013

### catherinenanc

Hm. That can be done, though, right? Finding it in terms of x alone? Since I think that is what my professor wants, how would I do that?

11. Mar 14, 2013

### Staff: Mentor

If that was what your professor wanted, why didn't he/she ask for that explicitly? There is nothing in the problem statement about solving for y' as a function of x alone.

In any case, having the variables occur in both the base and the exponent makes it very difficult to separate them. The only thing I can think of that would be helpful is the Lambert W function.

12. Mar 14, 2013

### catherinenanc

You're probably right, since this is a Calculus class and not anywhere near the level of a differential equations class, that's probably where we would stop.

Although, I guess I could simplify to:

dy/dx=[y[2x(ln y) - 5y]]/[x[5y(ln x) - 2x]]

13. Mar 14, 2013

Thanks!