<s>[INST] Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=\ln(-5(e^x+1)^5+c)$? This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$\frac{-e^x}{(e^x+1)^6}dx=\frac{e^y}{(e^y+1)^2}dy$$
$$\int{\frac{-e^x}{(e^x+1)^6}dx}=\int{\frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$\int{\frac{e^y}{(e^y+1)^2}dy}=\int{\frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$\int{u^{-2}dy}=-\int{v^{-6}}dx$$
$$-u^{-1} + c_1=\frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-\frac{1}{e^y+1}+c_1=\frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$\ln(-5(e^x+1)^5+c_3)=\ln(e^y)$$
$$\ln\left(-5(e^x+1)^5+c_3\right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
 [/INST] The wrong step is here :

$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
  Multiply out:
  $$-5(e^x+1)^5-c_3=e^y+1$$

The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$\left(-\frac{1}{e^y+1}\right)(e^y+1)\big(5(e^x+1)^5\big)=\left(\frac{1}{5(e^x+1)^5}+c_3\right)(e^y+1)\big(5(e^x+1)^5\big)$$
Simplify :
$$-\big(5(e^x+1)^5\big)=\left(1+c_3 5(e^x+1)^5\right)(e^y+1)$$
$$ e^y+1=\frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=\ln\left(\frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 \right)$$
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