Why can you Take out An E^xy??

bmed90

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

Im learning about implicit solutions for differential equations. Anyways I took the derivative of a relation containing x and y to get

dy/dy=1-e^xy(y)/ e^xy(x)+1


2. Relevant equations

Anywhoo it turns out to be a solution to the diff eq dy/dy = e^-xy - y/ e^-xy + x



3. The attempt at a solution

Apparently you can take out an e^xy from dy/dy=1-e^xy(y)/ e^xy(x)+1 in order to get to
dy/dy = e^-xy - y/ e^-xy + x

How exactly is this so? How does this work....I hope you understand my question

SammyS

In order for your question to be understood (without a lot of guessing by the reader) you really need to include parentheses where they're needed so that your mathematical expressions are unambiguous.

What is to be included in you numerators?

What is to be included in you denominators?

etc. ...

Joffan

dy/dy = 1. Always.

Assuming you mean dy/dx - I'll echo SammyS, who just posted while I was typing, and ask for more clarity. And it wouldn't hurt to post the original xy-relation, either.

bmed90

Alright So Basically I just want to know how one goes from this

dy/dy= 1-(e^xy)(y)/ (e^xy)(x)+1

to this

dy/dy = (e^-xy) - y/ (e^-xy) + x

by taking out an e^xy from the top and bottom

SammyS

Taking your parentheses literally you have:

dy/dy= 1-(e^xy)(y)/ (e^xy)(x)+1

which means:

[itex]\displaystyle \frac{dy}{dx}=1-(e^{xy})\frac{y}{e^{xy}}x+1\,.[/itex]

And the equation:

dy/dy = (e^-xy) - y/ (e^-xy) + x

which means:

[itex]\displaystyle \frac{dy}{dx}=(e^{-xy})-\frac{y}{e^{xy}}+x\,.[/itex]

On the other hand:

dy/dy= (1-e^(xy)(y)) / ((e^(xy))(x)+1)

means:

[itex]\displaystyle \frac{dy}{dx}=\frac{1-e^{xy}(y)}{e^{xy}(x)+1}\,.[/itex]