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? 
 A: hint: $x+y = (x+2) + (y-2)$ , and use $(A+B)^2 = A^2 + 2AB +B^2, A = x+2,B = y - 2$ to expand the numerator and simplify
A: If there is no typo in the equation, I do not think that a solution could be obtained.
The best I was able to do was, working with $x(y)$ instead of $y(x)$
$$x'=\frac{(y-2) (x+2)}{(x+y)^2}$$ Making $x=z-2$ gives
$$z'=\frac{(y-2) z}{(z+y-2)^2} $$ which leads to an implicit equation
$$\frac{z (3 \log (z)+\log (z+2 y-4)-1)-2 y+4}{4 z}=C$$ which looks impossible to solve.
A: 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 $k=\pm e^{-4c_1}$
A: $$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}=\left(\dfrac{x+y}{x+2}\right)^2\dfrac{x+2}{y-2}$$
let $w=\dfrac{x+y}{x+2}$ then $y=w(x+2)-x$ and
$$w'(x+2)+w-1=y'=w^2\dfrac{1}{w-1}$$
which is separable
$$\dfrac{w-1}{2w-1}dw=\dfrac{dx}{x+2}$$
with integration the solution will be obtained.
