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I have the below differential equation $$y' = \frac{y(y-2)}{x(y-1)}$$ and I solve it by cross multiplying, then integrating. The general solution that I get is $$y = 1 \pm \sqrt{1 + Cx^2}, ~ C\in \mathbb{R}.$$ Now, in class, my professor mentioned that if we are given the initial condition $y(0) = 0$, we have infinitely many solutions in the form of $$y = 1 - \sqrt{1 + Cx^2}, ~ C\in \mathbb{R}.$$ However, it seems to me that the original differential equation is undefined at our initial condition because we are dividing by zero. If the differential equation were given in the form of $$y'x(y - 1) = y(y-2),$$ then I would agree because we are not dividing by zero. Finally, my question is that given $y(0) = 0$, or $y(0) = 2$, are there solutions?

Thanks.

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    $\begingroup$ "However, it seems to me that the original differential equation is undefined at our initial condition because we are dividing by zero. " No, you should consider the limit $x\to0$. Example with $y(x)\simeq A x^2$ there's no dividing by $0$. $\endgroup$ – Dr. Wolfgang Hintze Feb 17 at 14:22
  • $\begingroup$ @Dr.WolfgangHintze How come i should consider the limit? $\endgroup$ – sepehr78 Feb 17 at 14:24
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    $\begingroup$ Because if you have a expression like $0/0$ which has no definite value, one should consider the vicinity of $x=0$. Look up the rule of l'Hospital. $\endgroup$ – Dr. Wolfgang Hintze Feb 17 at 14:31
  • $\begingroup$ L'hopital is for limits, but there is no limit here. $\endgroup$ – sepehr78 Feb 17 at 14:45
  • $\begingroup$ Again: if you have $0/0$ then this meaningless expression is given a meaning by applying the limiting procedure. BTW, you're dealing with a differential equation. Hence you already have a limit, that of the difference quotient, perhaps without noting it. Why not introduce another one? Perhaps it is better to ask your professor for a more precise definition of your task. Most surely you will understand better what I meant.. $\endgroup$ – Dr. Wolfgang Hintze Feb 17 at 16:02
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We're simply given the initial condition to find a value of $C$ to get the particular solution and notice that this initial condition is not to be plugged into the ODE but the function you get after solving it. In your case, every value of $C \in \mathbb{C}$ satisfies $y(0)=0$.

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  • $\begingroup$ But since we are finding a solution to the differential equation, it seems like the equation should make sense at our initial condition. $\endgroup$ – sepehr78 Feb 17 at 14:25
  • $\begingroup$ Have a look what's going on here: desmos.com/calculator/3qndakyjbs $\endgroup$ – Paras Khosla Feb 17 at 14:32
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$$y' = \frac{y(y-2)}{x(y-1)}$$ $$\frac{y-1}{y^2-2y}y' = \frac{1}{x}$$ $$\int\frac{y-1}{y^2-2y}dy = \int\frac{dx}{x}$$ $$\frac{1}{2}\ln{|y^2-2y|} = \ln{|x|}+C_1$$ $$y^2-2y=C_2x^2$$ $$y(0)=0\Rightarrow 0=C_2\cdot0\Rightarrow C_2 \in \mathbb{C}$$ $$y(0)=2\Rightarrow 0=C_2\cdot0\Rightarrow C_2 \in \mathbb{C}$$ $$\therefore y^2-2y=kx^2$$ For any $k\in\mathbb{C}$

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