<s>[INST] Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
 [/INST] i don't think you made a mistake. continuing from where you stopped,
if $(y+1)(x+1)>0$ , then $y+1=\frac{c_1}{x+1}$.
otherwise if $(y+1)(x+1)<0$ , then $y+1=-\frac{c_1}{x+1}$.
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