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
 A: 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.
A: From this step you use exponential
$$\ln\vert y+3\vert=-\ln\vert x^2-4\vert +C$$
$$\Rightarrow e^{\ln\vert y+3\vert}=e^{-\ln\vert x^2-4\vert +C}$$
$$\Rightarrow \vert y+3\vert=\frac{e^C}{e^{\ln\vert x^2-4\vert}}$$
$$\Rightarrow \vert y+3\vert=\frac{K}{\vert x^2-4\vert}$$
For some positive number $K$
$$\Rightarrow y+3=\frac{K}{x^2-4}\text{ or }y+3=-\frac{K}{x^2-4}$$
Which by allowing $K$ to be positive or negative can be consolidated to become
$$\Rightarrow y+3=\frac{K}{x^2-4}$$
$$\Rightarrow y=\frac{K}{x^2-4}-3; K\in\mathbb{R}.$$
Now the solution that you saw is equivalent to this one since when you take a common denominator you get
$$\Rightarrow y=\frac{-3x^2+12+K}{x^2-4}$$
And taking $12+K$ to be a constant
$C’$ you get
$$\Rightarrow y=\frac{-3x^2+C’}{x^2-4}$$
Which is the solution you saw in the answer sheet.
Hope this helps.
A: $$2xy+6x+(x^2-4)y'=0$$
It's easier to rewrite the DE as:
$$((x^2-4)y)'=-6x$$
Integrate:
$$(x^2-4)y=-3x^2+C$$
