Singular solution to ODE and singularities

Question

I'm a bit confused about those two, I'll try to explain. I get an ODE like $$ \frac {y \cdot dy }{\sqrt{y^2+1} } + \frac {x \cdot dx}{ \sqrt {x^2 +1}} = 0 $$ What I'm not sure about :

1) at $y=0$ the equation makes no sense right ? cause the coefficient of the derivative would be $0$

2) at $y= i = \sqrt{-1}$ , considering over $\mathbb{C}$ not $\mathbb{R}$ , there is not a constant solution but a singularity, right ? cause at this point the original equation is not well defined , right ? To have a constant solution would require a well defined equation condition to be satisfied first but the value $y=0$ is outside the domain, right ?

3) at $x= i = \sqrt{-1}$ again there's a singularity, cause we just can't divide by $0$

4) what if the equation is given as : $$ \frac {dy}{dx}=-\frac{x}{\sqrt{x^2+1}} \cdot \frac {\sqrt{y^2+1}}{y} $$ now we have a constant solution at $y=i=\sqrt{-1}$ , right ? cause this would make $$\frac {\sqrt{y^2+1}}{y}=0$$ and $$\frac{dy}{dx}=0$$ and the whole equation true ?

I'll add one more thing to get rid of the complex case : what if instead of $\sqrt{y^2+1}$ we had $\sqrt{y^2-1}$ and the same for x, assuming that's a good ODE that has a general solution, what happens at $y=1$ (real value now, not complex) ? Do we have a constant (probably singular) solution ? or is just a singularity ? I need to know what happens for $p(y)y'+q(x)y=0$ , at points when p(y) has singularities(like division by 0), is it a singular solution or just a singularity ? And if that changes if the equation is given as $y'=q(x) \cdot \frac{1}{p(y)}$ to start from.

Trying to reduce this further, sorry it's hard to be understood when you're (me) confused. Given 2 ODEs: $$\frac {y}{y-2}\cdot \frac{dy}{dx}=1$$ and $$\frac{dy}{dx}=1\cdot \frac{y-2}{y}$$ 1.They have the same solution set and they are the same.Right?

2.At $y=0$ the ODEs are impossible, undefined, they make no sense. Right ?

3.At $y=2$ they have a constant equilibrium solution , singular solution. Right?

Again, thanks for helping me.

Please help me with this confusion. Thank you.

Answer 1

Your first equation has general solution

$$\sqrt{y^2 + 1} + \sqrt{x^2 + 1} = C.$$

1. The equation is defined for varying $x$ and $y$. Since $y/\sqrt{y^2+1}$ is not identically $0$, the presence of $dy$ in this problem is necessary.

2. and 3. Both $x$ and $y$ are real variables, so you do not consider $x = i$ or $y = i$. Both $y/\sqrt{y^2 + 1}$ and $x/\sqrt{x^2 + 1}$ are continuous functions of $x$ and $y$.

The last part is similar to 2. and 3. The variables $x$ and $y$ are real, and the function $xy/(\sqrt{x^2 + 1}\sqrt{y^2 + 1})$ is a continuous function of $x$ and $y$.

Edit: Ok, I think I see where you're confused. A singular solution either has a singularity at some point or fails to satisfy uniqueness at all points. The equation $xy' + y = 0$ has singular solution $y = C/x$, where the singularity occurs at $x = 0$. One important example is the differential equation $y'^2 = 4y$. It has solutions $y = (x + C)^2$ and $y = 0$. The integral curves $y = (x + C)^2$ have common tangent line $y = 0$, so uniqueness of the zero solution fails and $y = 0$ is the singular solution.

Edit 2: Your questioning is unclear, but I hope this helps.

1. They are the same.

2. $y = 0$ is not a solution.

3. Yes.

Answer 2

I am not sure at all to address your concerns, so forgive me if I am off-topic.

The equation write $$\frac {y \cdot dy }{\sqrt{y^2+1} }=- \frac {x \cdot dx}{ \sqrt {x^2 +1}}$$ So, integration leads to $$\sqrt{y^2+1}=-\sqrt{x^2+1}+C$$ and so $$y=\pm\sqrt{c^2+x^2-2 c \sqrt{x^2+1}}$$ Is this helping for your analysis ?