If y (d y)/(d x)=x[y^2/x^2+(φ(y^2/x^2))/(φ^′(y^2/x^2))], x>0, φ>0, and y(1)=-1, then φ(y^2/4) is equal to:

If $\displaystyle y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x\left[\frac{y^2}{x^2}+\frac{\phi\left(\dfrac{y^2}{x^2}\right)}{\phi^{\prime}\left(\dfrac{y^2}{x^2}\right)}\right], x>0, \phi>0$, and $\displaystyle y(1)=-1$, then $\displaystyle \phi\left(\frac{y^2}{4}\right)$ is equal to :