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.
Best Answer
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 $y(0)=0$.