If the solution of the differential equation $\frac{dy}{dx}=\frac{(x+y{)}^2}{(x+2)(y-2)}\text{ is }(x+2{)}^m\left(1+\frac{k(y-2)}{x+2}\right)=C{e}^{2\frac{(y-2)}{x+2}}$ where $\mathrm{m},\mathrm{k}\in \mathrm{N}\text{ and }\mathrm{C}$ and C is arbitrary constant then $m+k$ equals.

$\frac{dy}{dx}=\frac{(x+y{)}^2}{(x+2)(y-2)}$

On putting $\mathrm{X}=\mathrm{x}+2\text{ and }\mathrm{Y}=\mathrm{y}-2$ , the given differential equation reduces to 

$\begin{array}{l}\frac{dY}{dX}=\frac{(X-2+Y+2)}{XY}=\frac{(X+Y{)}^2}{XY}\\ \text{ put }Y=VX\Rightarrow \frac{dY}{dX}=V+X\frac{dV}{dX}\\ \Rightarrow V+X\frac{dV}{dX}=\frac{(1+V{)}^2}{V}\\ \Rightarrow \frac{V}{2V+1}dV=\frac{dX}{X}\\ \Rightarrow \left(1-\frac{1}{1+2V}\right)dV=\frac{2dX}{X}\end{array}$

integrating, $\int \left(1-\frac{1}{1+2\mathrm{V}}\right)\mathrm{dV}=\int \frac{2\mathrm{dX}}{\mathrm{X}}$

$\begin{array}{l}\Rightarrow \mathrm{V}-\frac{1}{2}\ln (1+2\mathrm{V})=2\ln \mathrm{X}+\mathrm{C}\\ \Rightarrow {\mathrm{X}}^4\left(1+\frac{2\mathrm{Y}}{\mathrm{X}}\right)={\mathrm{Ke}}^{2\mathrm{Y}/\mathrm{X}}\end{array}$

where, $\mathrm{X}=\mathrm{x}+2,\text{ and }\mathrm{Y}=\mathrm{y}-2\text{ and }\mathrm{m}=4,\mathrm{k}=2$

$\therefore m+k=6$