Examine the singular points of the equation y'=(2y-x)/(3x+6)

Examine the singular points of the equation $$y'=\frac{2y-x}{3x+6}.$$

Step 1 · Symmetric form

Write the equation as $$P(x,y)\,dx - Q(x,y)\,dy = 0,$$ with $$P(x,y)=2y-x,\qquad Q(x,y)=3x+6=3(x+2).$$

Step 2 · Locating the singular point

A singular point requires $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 2y-x=0,\\ 3(x+2)=0. \end{cases}$$ From the second equation $x=-2$. Substituting into the first: $$2y-(-2)=0\implies 2y+2=0\implies y=-1.$$ The unique singular point is $$(x^*,y^*)=(-2,-1).$$

Step 3 · Shifting to the origin

Introduce new variables centred at the singular point: $$u=x+2,\qquad v=y+1,$$ so that $x=u-2$, $y=v-1$. Substituting: $$P=2y-x=2(v-1)-(u-2)=2v-u,\qquad Q=3(x+2)=3u.$$ The equation in the new variables is $$\frac{dv}{du}=\frac{2v-u}{3u},$$ with the singular point shifted to the origin $(u,v)=(0,0)$.

Step 4 · Associated linear autonomous system

Write the equivalent planar system $$\frac{du}{dt}=Q(u,v)=3u,\qquad \frac{dv}{dt}=P(u,v)=2v-u,$$ with coefficient matrix $$A=\begin{pmatrix}3&0\\-1&2\end{pmatrix}.$$

Step 5 · Eigenvalues of $A$

The matrix $A$ is lower triangular, so the eigenvalues are the diagonal entries: $$\lambda_1=3,\qquad \lambda_2=2.$$

Step 6 · Classification of the singular point

Both eigenvalues are real, distinct, and positive: $$0<\lambda_2=2<\lambda_1=3.$$ The origin (and hence the point $(-2,-1)$ in original coordinates) is an unstable node. All trajectories depart from the singular point as $t\to+\infty$.

Step 7 · Eigenvectors

For $\lambda_1=3$: $$(A-3I)\mathbf{v}=0 \implies \begin{pmatrix}0&0\\-1&-1\end{pmatrix}\mathbf{v}=0 \implies v_1+v_2=0,$$ so $\mathbf{v}^{(1)}=(1,-1)^\top$. In original coordinates this nodal axis is $y=-x-3$. For $\lambda_2=2$: $$(A-2I)\mathbf{v}=0 \implies \begin{pmatrix}1&0\\-1&0\end{pmatrix}\mathbf{v}=0 \implies v_1=0,$$ so $\mathbf{v}^{(2)}=(0,1)^\top$. In original coordinates this nodal axis is the vertical line $x=-2$.

Step 8 · General solution

$$\begin{pmatrix}u(t)\\v(t)\end{pmatrix} =C_1 e^{3t}\begin{pmatrix}1\\-1\end{pmatrix} +C_2 e^{2t}\begin{pmatrix}0\\1\end{pmatrix}.$$ Returning to original coordinates: $$x(t)=C_1 e^{3t}-2,\qquad y(t)=-C_1 e^{3t}+C_2 e^{2t}-1.$$

Step 9 · Family of integral curves

For $C_1\neq0$, eliminating $t$ gives $$(y+1)+(x+2)=K\,(x+2)^{2/3},\qquad K=\frac{C_2}{C_1^{2/3}},$$ i.e.\ in original coordinates $x+y+3=K\,(x+2)^{2/3}$. Special cases: $C_1=0$ gives the vertical line $x=-2$ (slower nodal axis), and $C_2=0$ gives the line $y=-x-3$ (faster nodal axis).

Step 10 · Behaviour of trajectories near the singular point

Since both eigenvalues are positive, all trajectories move away from $(-2,-1)$ as $t\to+\infty$. As $t\to-\infty$ every trajectory (except those on the faster axis $y=-x-3$) approaches the singular point tangentially to the slower eigendirection $x=-2$. This is the characteristic behaviour of an unstable node.

Step 11 · Conclusion

$$\boxed{ (x^*,y^*)=(-2,-1)\text{ is the unique singular point; it is an \textbf{unstable node}.} }$$ The eigenvalues $\lambda_1=3$ and $\lambda_2=2$ are both real, distinct, and positive. The two nodal axes in original coordinates are the line $y=-x-3$ (faster direction, $\lambda_1=3$) and the vertical line $x=-2$ (slower direction, $\lambda_2=2$). All trajectories depart from the singular point, approaching it tangentially to $x=-2$ as $t\to-\infty$.

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.