How are restrictions introduced to a differential equation addressed when we have found the equation's solutions? This is how I was shown how to solve the following differential equation:
$xy'=\sqrt{x^2-y^2}+y  \qquad \rightarrow \qquad y'=\sqrt{1-\frac{y^2}{x^2}}+\frac{y}{x} \qquad [1] $
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let $z=\frac{y}{x}\qquad \rightarrow y=zx \qquad \rightarrow \frac{dy}{dx}=z+x\frac{dz}{dx}=\sqrt{1-z^2}+z $
$\hspace{5.6cm}\rightarrow x\frac{dz}{dx}=\sqrt{1-z^2} $
$\hspace{5.6cm}\rightarrow \int \frac{1}{\sqrt{1-z^2}} dz=\int \frac{1}{x}dx \qquad$ provided $z^2\neq1 \hspace{0.7cm}and \hspace{0.6cm}x\neq0$
$\hspace{5.6cm}\rightarrow \arcsin(z)=\ln|x|+C$
$\hspace{5.6cm}\rightarrow \frac{y}{x}=\sin(\ln|Cx|)$
since $z=\pm 1 \hspace{0.5cm}$ was excluded from our workings we consider it as a possible solution and find out that it corresponds to a solution:
$y=\pm x$
I was told that dividing by x at step [1] was fine as the equation would be "rubbish" otherwise. I understand all the workings that follows after [1]; however, my not understanding of [1] I feel like points to a misconception I have of differential equations. In the next few lines I'll write my understanding of step [1] and would appreciate it if you'd correct me anywhere I'm wrong:
I think what my tutor meant when he said, the equation is useless when $x=0$ was when $x=0$ is being considered as a line and not as the coordinate of a point. It's reasonable to say that an expression containing $y'$ is useless when we're dealing with the line $x=0$ (or, as a matter of fact, any line of the form $x=a$).
Granted, that way of thinking about $x=0$ would make the differential equation useless for $x=0$. But, what about $x=0$ as the coordinate of a point? If it's just a point $y'$ no longer needs to be meaningless (as long as the point whose x coordinate is 0 is a point on a continuous curve). So, in other words, $x=0$ referring to a single point should be allowed and possibly correspond to points on the integral curves of the equation.
The solution $y=x$ to the equation suggests that $x=0$ makes sense as a coordinate; but I don't understand why. Why are we including points with x-coordinates $x=0$ in a solution that is dependent on a step at which we divide by x? $x=0$ should completely be excluded from any solutions that include in their derivation the step [1]. So, the solutions should instead be rewritten as:
$y=\pm x \quad$   where $x\neq 0 \quad$  and $ \quad \frac{y}{x}=\sin(\ln|Cx|) \quad$  where $\quad x \neq 0 \quad $   and   $\quad C \neq 0$
This would then give rise to the issue that all points with an x-coordinate $x=0$ are being completely ignored; while, they have as much right to be a part of the solution as any other point does (the fact that they're being excluded is simply because the only method that has yielded solutions uses the step [1]). So, how do we show that the point (0,0) does, in fact, belong to the integral curves $y=\pm x$?
I really hope that I've made at least some sense in the past few lines. I'd greatly appreciate it if you would correct my misconceptions about solving differential equations.
 A: A solution $y$ of the ODE is a function of $x$, so think of it as $y=y(x)$.
Your first step is to divide the ODE by $x$, which implies the condition $x\neq 0$ since it is not valid step if $x=0$. This only means that any solution at $x=0$ must be manually validated in the original equation, since it is necessarily excluded from that point on in your sequence of implications.
It isn't true that $x=0$ gives "rubbish" in the original equation. What is true is that when $x=0$, the original equation becomes $$0=\sqrt{-y^2} + y$$ and this is only meaningful if $y=0$ since the radicand would be negative otherwise.
In other words, it must be the case that $y(0) = 0$ if the solution $y(x)$ is to be defined at $x=0$. You must also determine whether $y'(0)$ exists, and if not, then state that $x=0$ is not a domain point of the solution $y(x)$.
A: At $x=0$ your equation loses the character of being a differential equation, as the coefficient of the only derivative becomes zero. In other words, for it to be an ODE, you have to exclude the line $x=0$ from the domain.
It may accidentally happen that solutions have limits for $x\to 0$ and, even more, that you can join pieces from both sides that give not only a continuous, but even a differentiable function. These compositions however are not solutions of the ODE in the "ordinary" sense, even though they satisfy the implicit differential equation through insertion.
