<s>[INST] The solution of differential equation $2xy+6x+(x^2-4)y'=0$ When I solved the above DE, I reached to the step and confused how to continue
$2x(y+3)+(x^2-4)dy/dx$
$(x^2-4)  dy/dx=-2x(y+3)$
$dy/(y+3)=-2x/(x^2-4) dx$
$∫dy/(y+3)=∫-2x/(x^2-4) dx$
$\ln⁡|y+3|=-ln⁡|x^2-4|+C$
$\ln⁡|y+3|+ln⁡|x^2-4|=C$
$\ln⁡|(y+3)(x^2-4)|=C$
$|(y+3)(x^2-4)|=e^C$
then how can  I get the function $y$ for the last equation and when I checked the solution by W.F, I found it  as
$y=c/(x^2-4)-3x^2/(x^2-4)$
plz can anyone help me to show how  happed that
thanks
 [/INST] Note from
$$ \ln⁡|(y+3)(x^2-4)|=C$$
you have
$$ y=-3+\frac{C}{x^2-4}=\frac{-3x^2+12+C}{x^2-4}=\frac{-3x^2}{x^2-4}+\frac{12+C}{x^2-4} $$
Now define $C_1=12+C$ and you have
$$ y=\frac{-3x^2}{x^2-4}+\frac{C_1}{x^2-4} $$
which is the same as the answer.
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