Singular solution of differential eqn

Question

I want to find singular solutions of $dy/dx = x\sqrt{1-y^2}$ and $\left(e^x+e^{-x}\right)dy/dx = y^2$. For the first one when I solved the differential equation I got $y= \sin(x^2/2+c)$ so that would exclude $y = 1$ or $-1$ as singular solutions right? So what would the singular solution be? And for the second one, is the singular solution simply $y =0$?

Answer 1

$$\frac{dy}{dx}= x\sqrt{1-y^2}\tag 1$$ Obviously the functions $\quad y(x)=1\quad$ and $\quad y(x)=-1\quad$ are solutions of Eq.$(1)$. Observe that they don't belong to the general solution : $\quad y(x)=\sin\left(\frac 12 x^2-C\right)\quad$ for all values of $x\in \Re$.

Of course, for example for $x=\sqrt{2C+\pi}$ we have $y(x_0)=1$. But this is only a point $(x_0,1)$, not the whole function $y(x)=1$. So, the function $y(x)=1$ doesn't belong to the above general solution.

$$(e^x+e^{-x})\frac{dy}{dx}= y^2\tag 2$$ Obviously the function $\quad y(x)=0\quad$ is solution of Eq.$(2)$. Observe that this solution doesn't belong to the general solution : $\quad y(x)=\frac{1}{C-\tan^{-1}(e^x)}\quad$ for all values of $x\in \Re$.

Of course, for $x_0=\ln|\tan(C)|$ we have $y(x_0)=0$. But this is only a point $(x_0,0)$, not the whole function $y(x)=0$. So, the function $y(x)=0$ doesn't belong to the above general solution.