<s>[INST] Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ 
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by: 
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d) None of these

Attempt: 
I have expanded and checked but couldn't spot any exact differentials. 
Secondly, it's not a homogeneous equation, so couldn't use $y = vx$. 
How do I go about solving this problem? 
 [/INST] Let $X=x+2$ and $Y=y-2$, so the given DE is equivalent to
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}$$
last DE is homogeneous, so can be transformed into a separable DE by making $Y =uX$ as follows
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}\qquad\iff\qquad u+X\frac{\mathrm d u}{\mathrm dX}=\frac{X^2(1+u)^2}{X^2u}$$
Then we get
$$X\frac{\mathrm d u}{\mathrm d X}=\frac{1+2u}{u}\quad\implies\quad \frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}=\frac1X$$
Integrating both sides respect to $X$ (supposing that $u$ depends on $X$) we get
\begin{align*}
\int\frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}\mathrm d X&=\int\frac1X\mathrm dX\\[4pt]
\int\left(\frac12-\frac{1/2}{2u+1}\right)\mathrm du&=\int\frac1X\mathrm dX\\[4pt]
\frac12u-\frac14\ln|2u+1|&=\ln|X|+c_1
\end{align*}
Last equality is equivalent to
$$\frac{y-2}{x+2}-\frac12\ln\left|\frac{2(y-2)}{x+2}+1\right|=2\ln|x+2|+2c_1$$
Notice that this solution can be carried to the form b):
\begin{align*}
\frac{2(y-2)}{x+2}-4c_2&=4\ln|x+2|+\ln\left|\frac{2(y-2)}{x+2}+1\right|\\[4pt]
k e^{\frac{2(y-2)}{x+2}}&=(x+4)^4\left[\frac{2(y-2)}{x+2}+1\right]
\end{align*}
where $k=\pm e^{-4c_1}$
</s>