Examine the singular points of the equation…

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

Step 1 · Symmetric form

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

Step 2 · Locating the singular points

From $P=0$: $y=x$ or $y=-x$. From $Q=0$: $x=1$ or $y=2$. Checking all combinations gives four singular points: $$(1,1),\quad(2,2),\quad(1,-1),\quad(-2,2).$$

Step 3 · Linearisation — general setup

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

Step 4 · Analysis at $(1,1)$

$J(1,1)=\begin{pmatrix}-2&0\\-2&2\end{pmatrix}$ is lower triangular, so $\lambda_1=-2$, $\lambda_2=2$: real, opposite sign, so $(1,1)$ is a saddle point. Eigenvector for $\lambda_1=-2$ is $(2,1)^\top$ (stable separatrix $2y=x+1$); for $\lambda_2=2$ it is $(0,1)^\top$ (unstable separatrix $x=1$).

Step 5 · Analysis at $(2,2)$

$J(2,2)=\begin{pmatrix}0&2\\-4&4\end{pmatrix}$, characteristic equation $\lambda^2-4\lambda+8=0$, giving $\lambda_{1,2}=2\pm2i$: complex with positive real part, so $(2,2)$ is an unstable focus. Evaluating $\dot y$ at $(3,2)$ gives $-5<0$, so the spiral is clockwise.

Step 6 · Analysis at $(1,-1)$

$J(1,-1)=\begin{pmatrix}-6&0\\-2&-2\end{pmatrix}$ is lower triangular, so $\lambda_1=-6$, $\lambda_2=-2$: real, distinct, negative, so $(1,-1)$ is a stable node. Eigenvector for $\lambda_1=-6$ is $(2,1)^\top$ (faster direction, $2y=x-3$); for $\lambda_2=-2$ it is $(0,1)^\top$ (slower direction, $x=1$). All trajectories approach $(1,-1)$ tangentially to $x=1$ as $t\to+\infty$.

Step 7 · Analysis at $(-2,2)$

$J(-2,2)=\begin{pmatrix}0&-6\\4&4\end{pmatrix}$, characteristic equation $\lambda^2-4\lambda+24=0$, giving $\lambda_{1,2}=2\pm2\sqrt5\,i$: complex with positive real part, so $(-2,2)$ is an unstable focus. Evaluating $\dot y$ at $(-1,2)$ gives $3>0$, so the spiral is counter-clockwise.

Step 8 · Summary

Step 9 · Conclusion

$$\boxed{ \begin{aligned} &(1,\phantom{-}1):\;\textbf{saddle},\quad \lambda_{1,2}=-2,+2.\\ &(2,\phantom{-}2):\;\textbf{unstable focus},\quad \lambda_{1,2}=2\pm2i\text{ (clockwise)}.\\ &(1,-1):\;\textbf{stable node},\quad \lambda_{1,2}=-6,-2.\\ &(-2,\phantom{-}2):\;\textbf{unstable focus},\quad\lambda_{1,2}=2\pm2\sqrt{5}\,i\text{ (counter-clockwise)}. \end{aligned} }$$

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.