If ydydx=x[y^2x^2+(y^2x^2)]!/'!(y^2x^2), x0, 0, y(1)=-1, then (y^24) is:

Question

If $y\dfrac{dy}{dx}=x\left[\dfrac{y^2}{x^2}+\phi\left(\dfrac{y^2}{x^2}\right)\right]\!\Big/\phi'\!\left(\dfrac{y^2}{x^2}\right)$, $x>0$, $\phi>0$, $y(1)=-1$, then $\phi\left(\dfrac{y^2}{4}\right)$ is:

Accepted Answer

With $t=y/x$, the equation reduces to $\dfrac{t\phi'(t^2)}{\phi(t^2)}\,dt=\dfrac{dx}{x}$; integrating with $u=t^2=y^2/x^2$:
$$\frac12\ln\phi(u)=\ln x+C.$$

Using $(1,-1)$: $C=\dfrac12\ln\phi(1)$. At $x=2$ (i.e. $u=y^2/4$): $\phi\left(\dfrac{y^2}{4}\right)=4\phi(1)$.

Solutions · 1