Examine the singular points of the equation…

Examine the singular points of the equation $$y'=\frac{2x}{1-x^2-y^2}.$$

Step 1 · Symmetric form

Write the equation as $$P(x,y)\,dx - Q(x,y)\,dy = 0,$$ with $$P(x,y)=2x,\qquad Q(x,y)=1-x^2-y^2.$$

Step 2 · Locating the singular points

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 2x=0,\\ 1-x^2-y^2=0. \end{cases}$$ From the first equation $x=0$. Substituting into the second: $y^2=1$, so $y=\pm1$. There are two singular points: $$(x_1^*,y_1^*)=(0,1)\qquad\text{and}\qquad(x_2^*,y_2^*)=(0,-1).$$

Step 3 · Linearisation — general setup

The associated planar system is $$\frac{dx}{dt}=Q(x,y)=1-x^2-y^2,\qquad \frac{dy}{dt}=P(x,y)=2x,$$ and the Jacobian matrix is $$J(x,y)=\begin{pmatrix}-2x & -2y \\ 2 & 0\end{pmatrix}.$$

Step 4 · Analysis at $(0,1)$

The Jacobian is $J(0,1)=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$, with eigenvalues $\lambda_{1,2}=\pm2i$ (purely imaginary). For a linear system this would be a centre; since the system is nonlinear, the linearisation alone is inconclusive and a first integral is needed to decide between a centre and a focus.

Step 5 · Analysis at $(0,-1)$

The Jacobian is $J(0,-1)=\begin{pmatrix}0&2\\2&0\end{pmatrix}$, with characteristic equation $\lambda^2-4=0$, so $\lambda_{1,2}=\pm2$ (real, opposite sign). The singular point $(0,-1)$ is a saddle point. For $\lambda_1=-2$: eigenvector $(1,-1)^\top$, stable separatrix tangent to $y=-x-1$. For $\lambda_2=2$: eigenvector $(1,1)^\top$, unstable separatrix tangent to $y=x-1$.

Step 6 · Exact solution and nature of $(0,1)$

Rewriting the equation as $2x\,dx/dy=1-x^2-y^2$ and substituting $w=x^2$ gives the linear ODE $w'+w=1-y^2$ (in $y$). Its integrating factor is $e^y$, and solving gives $$x^2+(y-1)^2=Ce^{-y},\quad C\in\mathbb{R}.$$ To confirm whether $(0,1)$ is a centre or a focus, define $F(x,y)=\bigl[x^2+(y-1)^2\bigr]e^{y}$. Differentiating along trajectories and substituting $\dot x=1-x^2-y^2$, $\dot y=2x$ shows $\dot F\equiv0$ identically, so $F$ is a genuine first integral. Every trajectory lies on a level curve $F=\text{const}$, which for small positive $C$ are closed curves surrounding $(0,1)$. Therefore $(0,1)$ is a true centre.

Step 7 · Summary of both singular points

Step 8 · Conclusion

$$\boxed{ \begin{aligned} &(0,1):\;\text{\textbf{centre}} \text{ — trajectories are closed curves } x^2+(y-1)^2=Ce^{-y},\; C>0.\\ &(0,-1):\;\text{\textbf{saddle point}} \text{ — eigenvalues }\lambda_{1,2}=\pm2, \text{ separatrices tangent to }y=\pm x-1. \end{aligned} }$$ The first integral $F=\bigl[x^2+(y-1)^2\bigr]e^{y}=\mathrm{const}$ confirms that all trajectories near $(0,1)$ are closed, establishing it as a true centre of the nonlinear system, not merely of its linearisation.

Summary Table

Additional Notes

• Each singular point is classified independently via the Jacobian of the associated planar system evaluated at that point; the overall phase portrait combines all of them.