Examine the singular points of the equation…

Examine the singular points of the equation $$y'=\frac{x(2y-x+5)}{x^2+y^2-6x-8y}.$$

Step 1 · Symmetric form

Write the equation as $$P(x,y)\,dx - Q(x,y)\,dy = 0,$$ with $$P(x,y)=x(2y-x+5),\qquad Q(x,y)=x^2+y^2-6x-8y.$$

Step 2 · Locating the singular points

From $P=0$: $x=0$ or $2y-x+5=0$. Completing the square on $Q=0$ gives $(x-3)^2+(y-4)^2=25$, a circle. Solving both cases against this circle gives four singular points: $$(0,0),\quad(0,8),\quad(7,1),\quad(3,-1).$$

Step 3 · Linearisation — general setup

The associated planar system is $$\frac{dx}{dt}=Q(x,y)=x^2+y^2-6x-8y,\qquad \frac{dy}{dt}=P(x,y)=x(2y-x+5),$$ with Jacobian $$J(x,y)=\begin{pmatrix}2x-6 & 2y-8\\2y-2x+5 & 2x\end{pmatrix}.$$

Step 4 · Analysis at $(0,0)$

$J(0,0)=\begin{pmatrix}-6&-8\\5&0\end{pmatrix}$, characteristic equation $\lambda^2+6\lambda+40=0$, giving $\lambda_{1,2}=-3\pm\sqrt{31}\,i$: complex with negative real part, so $(0,0)$ is a stable focus. Evaluating $\dot y$ at $(1,0)$ gives $4>0$, so the spiral is counter-clockwise.

Step 5 · Analysis at $(0,8)$

$J(0,8)=\begin{pmatrix}-6&8\\21&0\end{pmatrix}$, characteristic equation $\lambda^2+6\lambda-168=0$, giving $\lambda_{1,2}=-3\pm\sqrt{177}\approx -16.304,\,10.304$: real, opposite sign, so $(0,8)$ is a saddle point.

Step 6 · Analysis at $(7,1)$

$J(7,1)=\begin{pmatrix}8&-6\\-7&14\end{pmatrix}$, characteristic equation $\lambda^2-22\lambda+70=0$, giving $\lambda_{1,2}=11\pm\sqrt{51}\approx 3.859,\,18.141$: real, distinct, positive, so $(7,1)$ is an unstable node.

Step 7 · Analysis at $(3,-1)$

$J(3,-1)=\begin{pmatrix}0&-10\\-3&6\end{pmatrix}$, characteristic equation $\lambda^2-6\lambda-30=0$, giving $\lambda_{1,2}=3\pm\sqrt{39}\approx -3.245,\,9.245$: real, opposite sign, so $(3,-1)$ is a saddle point.

Step 8 · Summary

Step 9 · Conclusion

$$\boxed{ \begin{aligned} &(0,0):\;\textbf{stable focus},\quad\lambda_{1,2}=-3\pm\sqrt{31}\,i \quad\text{(counter-clockwise spiral, asymptotically stable)}.\\[4pt] &(0,8):\;\textbf{saddle point},\quad\lambda_{1,2}=-3\pm\sqrt{177}.\\[4pt] &(7,1):\;\textbf{unstable node},\quad\lambda_{1,2}=11\pm\sqrt{51}.\\[4pt] &(3,-1):\;\textbf{saddle point},\quad\lambda_{1,2}=3\pm\sqrt{39}. \end{aligned} }$$

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.