Checking an implicit solution of a diff eqn

[df606]

Homework Statement

"Determine whether the equations on the right define implicit functions of x. For those which do, determine whether they are implicit solutions of the differential equations on the left."

e^(x-y) + e^(y-x)(dy/dx) = 0, e^(2y)+e^(2x) = 1

The Attempt at a Solution

Apologies for blurry camera picture, I still don't know latex.
I solved the right equation for y, then took the derivative, and plugged both back into the left equation. I've been looking at this for about half an hour and I can't figure out where I've gone wrong.

edit: the answer says that it is an implicit solution

 

[Dick]

You should probably try using implicit differentiation, like your title suggests. If e^(2x)+e^(2y)=1 then 2*e^(2x)+2*y'*e^(2y)=0 differentiating both sides by x, right? Solve for y' and substitute into the differential equation. It's often a lot easier then solving for y (which might not even be possible) and then differentiating.

 

[micromass]

Hi df606!

You wrote something like this two times:

e^{\frac{1}{2}\log(1-e^{2x})}=\sqrt{e}(1-e^{2x})

You seem to be doing e^{ab}=e^ae^b, which is of course incorrect. The correct formula is

e^{\frac{1}{2}\log(1-e^{2x})}=\left( e^{\log(1-e^{2x})}\right)^{1/2}=\sqrt{1-e^{2x}}

I did not check from there on...

 

[df606]

Aha! Thank you micromass. I figured I was doing something like that. However, after doing what Dick suggested, the problem became incredibly simple, and I find it rather hilarious that I didn't try doing that earlier.