Find y' if x[SUP]5y[/SUP]=y[SUP]2x[/SUP]

[catherinenanc]

Homework Statement

Find y' if x^5y=y^2x.

Homework Equations

d(ln y)=dy/y.

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!

 

[Mark44]

Homework Statement

Find y' if x^5y=y^2x.

Homework Equations

d(ln y)=dy/y.

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!

What do you get when you take the log (natural) of each side?
IOW, what is ln(x^5y)? What is ln(y^2x)?

 

[catherinenanc]

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

 

[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).

 

[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.

 

[Mark44]

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

Right.

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).

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

 

[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]

 

[catherinenanc]

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

 

[Mark44]

(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]

This looks fine.

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

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

 

[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?

 

[Mark44]

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.

 

[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]]

 

[catherinenanc]

Thanks!