I am trying to find the regions for the differential equation $\frac{dy}{dx}=\sqrt{xy} $ for which the solution is unique, with graph passing through the point $(x_0, y_0)$ in these regions.

I tried to applied Picard theorem for existence and uniqueness. The function $\frac{\partial f}{\partial y}(x,y) = \frac{\sqrt x}{2 \sqrt y}$ is defined when $xy\geq 0$ and $y \neq 0$ thus we have existence and uniqueness in the regions $R = [0, \infty)\times (0, \infty)$ and $R = (-\infty, 0]\times (-\infty, 0)$ for any $(x_0, y_0) \in R$.

However, answer is $R = (0, \infty)\times (0, \infty)$ and $R = (-\infty, 0)\times (-\infty, 0)$.

My question is why not $x=0$ is included in the region of existence and uniquness?

Thank you

  • 1
    $\begingroup$ That's because $y(x)=\frac19x^3$ and $y(x)=0$ are two different solutions with $y(0)=0.$ $\endgroup$ – Professor Vector Aug 5 '17 at 7:52
  • $\begingroup$ @ProfessorVector Thanks for the comment. Could you right the answer with more details? $\endgroup$ – mathscrazy Aug 8 '17 at 23:07

Your analysis is correct. The only reason to exclude $x=0$ would be that the requested regions have to be open sets by the definition of the term "region" or "domain of the ODE".

This could be also be implied in

"graph passing through the point $(x_0,y_0)$ in these regions"

because to "pass through" there needs to be a left and right of $x_0$, resp. $x_0$ needs to be an inner point of the domain of the solution, which is not given if $x_0=0$

  • $\begingroup$ Thank you for the answer. I understood completely. $\endgroup$ – mathscrazy Aug 13 '17 at 3:47

The function $y(x)=0$ for all $x$ is a solution. So from now on we consider only functions with $y(x)\neq 0$ for some $x$. For points with $y(x)\neq 0$ we have \begin{equation}\sqrt{x}=\frac{y'}{\sqrt{y}}=(2\sqrt{y})', \end{equation} so integrating both sides yields \begin{equation}\frac{2}{3}x^{\frac{3}{2}}+C=2\sqrt{y}, \end{equation} which impliess \begin{equation}y=\left(\frac{1}{3}x^{\frac{3}{2}}+c \right)^2. \end{equation}

  • $\begingroup$ Thanks a lot for the analysis. $\endgroup$ – mathscrazy Aug 13 '17 at 3:46

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