Differential Equations

$(x^2+ xy + 4x + 2y + 4)\frac{dy}{dx}-y^2=0$$((x+2)^2+y(x+2))\frac{dy}{dx}=y^2$Let $x+2=X,y=Y$$(X)(X+Y)\frac{dY}{dX}=Y^2$$-X^2dY=XYdY-Y^2dX$$-X^2dY=Y(XdY-YdX)$$-\frac{dY}{Y}=\frac{XdY-YdX}{X^2}$$-\ln|Y|=\frac{Y}{X}+C$$-\ln|y|=\frac{y}{x+2}+C$∵ it is passing through$(1,3)-\ln3=1+C$$C+-1-\ln3$∵ curve $\frac{y}{x+2}=\ln|y|-1-\ln3=0,x>0$......(i)put y=x+2 in equation (i) then $\frac{x+2}{x+2}+\ln|x+2|-1-\ln3=0$$x=1,-5$(reject)∴ curve intersect $y=x=2$ at point $(1,3)$for option (C),put $y=(x+2)^2$,we will get$x+2+2\ln(x+2)=1=\ln3$Clearly left hand side is an increasing functionHence, it is always greater than $2+2\ln2$ therefore no solutionfor option (C) put $y=(x+3)^2$ in equation (i)$\frac{(x+3)^2}{x+2}+\ln((x+3)^2-1-\ln3=0$$\frac{(x+3)^2}{x+2}+\frac{\ln(x+3)^2}{3}-1=0$$\because x>0 \Rightarrow x=3>x+2$                     and $x+3>3$So $\frac{(x+3)^2}{x+2}+\frac{\ln(x+3)^2}{3}>1$$\therefore \frac{(x+3)^2}{x+2}+\frac{\ln(x+3)^2}{3}-1=0$⇒curve $y=(x+3)^2$ does not intersect