Recent content by catherinenanc

Thanks!

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]]

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?

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

(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]

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.

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).

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

Homework Statement Find y' if x5y=y2x. 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...

1. Can anybody elxplain to me (or point me to a URL of an explanation) how Cantor proved the existence of a bijection (0,1]~[0,1]? 2. It's not for homework. I have to understand it generally for a paper I am writing. 3. I think it has something to do with transfinites but I can't get it.

Does this apply to the problem?

Ok, thanks!

So I show that xy1x-1=xy2x-1→y1=y2 by simply left-multiplyng both sides by x-1 and right-multiplying both sides by x? Is that too simple? Also, does my thinking on surjective work?

Ok, I am seeing an example and a counterexample. f={(1,1),(2,2)} and g={(1,1),(2,2),(3,2)}. Then f is one-to-one, g is not, and g°f={(1,1),(2,2)} is. Right? This works because the domain of g is not restricted to the image of f. Also, f={(1,1),(2,2),(3,2)} and g={(1,1),(2,2)}. Then f...

Also, the next problem is: "Let f:A→B and g:B→C be maps such that g°f is injective. Prove that f must be injective." So, twice he's said that it is f that must be injective. I just can't figure out why g doesn't have to be...