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Why can you Take out An E^xy?by bmed90
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#1
Jan1912, 03:55 PM

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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=1e^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=1e^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 


#2
Jan1912, 04:02 PM

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What is to be included in you numerators? What is to be included in you denominators? etc. ... 


#3
Jan1912, 04:08 PM

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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 xyrelation, either. 


#4
Jan1912, 04:22 PM

P: 99

Why can you Take out An E^xy?
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 


#5
Jan1912, 04:41 PM

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


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