Examine the singular points of the equation y'=(2x+y)/(x-2y-5)

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

Step 1 · Symmetric form

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

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 2x+y=0,\\ x-2y-5=0. \end{cases}$$ From the first equation $y=-2x$. Substituting into the second: $$x-2(-2x)-5=0\implies x+4x-5=0\implies 5x=5\implies x=1,\quad y=-2.$$ The unique singular point is $$(x^*,y^*)=(1,-2).$$

Step 3 · Shifting to the origin

Introduce new variables $u=x-1$, $v=y+2$, so $x=u+1$, $y=v-2$. Substituting: $$P=2(u+1)+(v-2)=2u+v,\qquad Q=(u+1)-2(v-2)-5=u-2v.$$ The equation in the shifted variables is $\dfrac{dv}{du}=\dfrac{2u+v}{u-2v}$, with singular point now at the origin $(u,v)=(0,0)$.

Step 4 · Associated linear autonomous system

Write the equivalent planar system: $$\frac{du}{dt}=Q(u,v)=u-2v,\qquad \frac{dv}{dt}=P(u,v)=2u+v,$$ with coefficient matrix $$A=\begin{pmatrix}1&-2\\2&1\end{pmatrix}.$$

Step 5 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (1-\lambda)^2+4=0 \implies \lambda^2-2\lambda+5=0.$$ Using the quadratic formula: $$\lambda=\frac{2\pm\sqrt{4-20}}{2}=1\pm 2i.$$ Thus $$\lambda_{1,2}=1\pm 2i.$$

Step 6 · Classification of the singular point

The eigenvalues are complex with positive real part: $$\operatorname{Re}(\lambda_{1,2})=1>0,\qquad \operatorname{Im}(\lambda_{1,2})=\pm2\neq0.$$ The origin (and hence $(1,-2)$ in original coordinates) is an unstable focus (spiral source). Trajectories spiral away from the singular point as $t\to+\infty$.

Step 7 · Direction of rotation

Evaluating $\dot{v}=2u+v$ at $(u,v)=(1,0)$ gives $\dot v=2>0$, so the trajectory moves upward on the positive $u$-axis, indicating a counter-clockwise spiral.

Step 8 · Eigenvector and general real solution

For $\lambda_1=1+2i$: $(A-\lambda_1 I)\mathbf v=0$ gives $\mathbf v^{(1)}=(1,-i)^\top=(1,0)^\top+i(0,-1)^\top$. With $\alpha=1$, $\beta=2$, $\mathbf p=(1,0)^\top$, $\mathbf q=(0,-1)^\top$: $$u(t)=e^{t}(C_1\cos 2t+C_2\sin 2t),\qquad v(t)=e^{t}(C_1\sin 2t-C_2\cos 2t).$$ Returning to original coordinates $x=u+1$, $y=v-2$: $$x(t)=e^{t}(C_1\cos 2t+C_2\sin 2t)+1,\qquad y(t)=e^{t}(C_1\sin 2t-C_2\cos 2t)-2.$$

Step 9 · Behaviour of trajectories near the singular point

The distance from the singular point is $r(t)=e^{t}\sqrt{C_1^2+C_2^2}$, which grows exponentially as $t\to+\infty$ (confirming instability) and tends to zero as $t\to-\infty$, so every trajectory approaches $(1,-2)$ asymptotically while performing infinitely many counter-clockwise revolutions with angular frequency $\omega=2$ (period $T=\pi$).

Step 10 · Conclusion

$$\boxed{ (x^*,y^*)=(1,-2)\text{ is the unique singular point; it is an \textbf{unstable focus}.} }$$ The eigenvalues $\lambda_{1,2}=1\pm2i$ are complex with positive real part $\operatorname{Re}(\lambda)=1>0$ and non-zero imaginary part $\operatorname{Im}(\lambda)=\pm2$. All trajectories spiral counter-clockwise away from $(1,-2)$ with angular frequency $\omega=2$ and exponential growth rate $\alpha=1$, confirming the unstable focus classification.

Summary Table

Additional Notes

• The classification follows from the standard theory of planar linear (or linearised) autonomous systems, based on the sign and nature of the eigenvalues of the coefficient/Jacobian matrix.