# Thread: One more DE question

1. ## One more DE question

Im a bit hazy on my log rules....and this problem isnt helping...

Original problem: ((e^y+1)^2)e^-ydx+ ((e^x+1)^3)e^-xdy=0

Where i've gotten: -((e^x+1)^-3)e^xdx= ((e^y+1)^-2))e^y dy

im not sure how to get rid of those e^x and the e^y.....probaly some log rule im missing, Thanks for any help!

2. Hello, neven87!

Problem: . $(e^y+1)^2e^{-y}dx + (e^x+1)^3e^{-x}dy\:=\:0$

Answer: . $(e^x+1)^{-2} + 2(e^y+1)^{-1}\:=\:C$

Re-arrange the equation: . $\frac{(e^y + 1)^2}{e^y}\ dx + \frac{(e^x + 1)^3}{e^x}\ dy \:=\:0\quad\Rightarrow\quad\frac{(e^x + 1)^3}{e^x}\,dy \;=\;-\frac{(e^y + 1)^2}{e^y}$

Separate variables: . $\frac{e^y}{(e^y+1)^2}\,dy \;=\;-\frac{e^x}{(e^x+1)^3}\,dx\quad\Rightarrow\quad (e^y+1)^{-2}(e^y\,dy) \;=\;-(e^x+1)^{-3}(e^x\,dx)$

Integrate: . $\int(e^y+1)^{-2}(e^y\,dy) \;=\;-\int(e^x+1)^{-3}(e^x\,dx)$

Let $u = e^y + 1\quad\Rightarrow\quad du = e^y\,dy$
Let $v = e^x + 1\quad\Rightarrow\quad dv= e^x\,dx$

Substitute: . $\int u^{-2}du \;=\;-\int v^{-3}\,dv\quad\Rightarrow\quad -u^{-1}\;=\;\frac{v^{-2}}{2} + c$

. . $\frac{v^{-2}}{2} + u^{-1}\;=\;c\quad\Rightarrow\quad v^{-2} + 2u^{-1} \;=\;C$

Back-substitute: . $(e^x + 1)^{-2} + 2(e^y + 1)^{-1} \;=\;C$