Can a point be in the solution curve of a differential equation when domain of that differential equation does not contain it? I am given to solve this : $\frac{dy}{dx}= \frac{2xy}{x^2-y^2}$ with initial condition $y(1) =1$
$(1,1)$ does not belong to the domain of this O.D.E : $\frac{dy}{dx}= \frac{2xy}{x^2-y^2}$
I have a very little knowledge of Differential equation. I can not understand how $(1,1)$ is belonging to the solution curve when domain of O.D,E does not contain it.
Can anyone please help me ?
 A: You're right that the question is meaningless as it stands. But you could reformulate the ODE as $(x^2-y^2) \frac{dy}{dx} = 2xy$, or the initial condition as $\lim\limits_{x \to 1} y(x) = 1$, and then the problem would make sense.
A: The source of the confusion here is that the phrase "finding a solution" is completely ambiguous. In most contexts, this is not an issue, but it is an issue when using functions that involve singularities. The careful, rigorous way to go about this is to give me a family of functions you are interested in. I would assume you are interested in functions of the family $y:\mathbb{R}\rightarrow\mathbb{R},$ and furthermore, I would assume that you want functions that are differentiable everywhere, a.k.a, differentiable at every point in their domain, which in this case, would be every point in $\mathbb{R}.$ With those assumptions in place, it now makes perfect sense to ask about solutions to $$y'=\frac{2xy}{x^2-y^2},\,y(1)=1.$$ However, with those specific assumptions in place, you would be correct in saying that the initial condition is nonsensical.
With that being said, as Hans Lundmark clarified in his answer, it is entirely possible that the initial condition as provided is actually an abuse of notation, and they probably meant to say that $$\lim_{x\to1}y(x)=1$$ instead. This amounts to saying that the intended solution should have a removable singularity or a branch singularity at $1.$ I should say, though, that rewriting the equation as $[x^2-y(x)^2]y'(x)=2xy(x)$ would not actually solve the problem, because for $y(1)=1,$ this rewrite implies $0y'(1)=2,$ which is nonsensical. So it still would be the case that the solution would not actually be a solution, since it would not be differentiable at $1.$ This is why it makes more sense to think of it as the solution actually having a singularity at $1.$ Hence, if instead, we assume that the class of functions you are interested in are the functions of the family $y:\mathbb{R}\setminus\{1\}\rightarrow\mathbb{R}$ instead, such that it is differentiable everywhere, with the rewritten boundary condition, then the problem to solve actually makes sense, and the equation can then be solved, though the solving method is annoying.
