Examine the singular points of the system dotx=3x-2y…

Examine the singular points of the system $$\dot{x}=3x-2y,\qquad \dot{y}=4y-6x.$$

Step 1 · Locating the singular point

Set both right-hand sides equal to zero: $$\begin{cases} 3x-2y=0,\\ -6x+4y=0. \end{cases}$$ The second equation is exactly $-2$ times the first, so the system reduces to the single condition $3x-2y=0$, i.e.\ $y=\tfrac{3}{2}x$. This means there is a whole line of singular points $$y=\tfrac{3}{2}x,\quad x\in\mathbb{R},$$ not just an isolated point. In particular $(0,0)$ is a singular point, and so is every point of the form $(2s,3s)$, $s\in\mathbb{R}$.

Step 2 · Coefficient matrix and its rank

The coefficient matrix is $$A=\begin{pmatrix}3&-2\\-6&4\end{pmatrix}.$$ Its determinant is $$\det A = 3\cdot4-(-2)(-6)=12-12=0,$$ confirming that $A$ is singular and that the system has a non-isolated set of equilibria (the line $y=\tfrac{3}{2}x$).

Step 3 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (3-\lambda)(4-\lambda)-(-2)(-6)=0 \implies \lambda^2-7\lambda=0 \implies \lambda(\lambda-7)=0.$$ Thus $$\lambda_1=0,\qquad \lambda_2=7.$$

Step 4 · Classification of the singular point

One eigenvalue is zero and the other is positive: $$\lambda_1=0,\qquad \lambda_2=7>0.$$ The zero eigenvalue reflects the line of equilibria ($\det A=0$). The non-zero eigenvalue $\lambda_2=7>0$ governs motion transverse to the line of equilibria. This configuration is called a non-isolated (degenerate) unstable node: the origin (and every point on $y=\tfrac{3}{2}x$) is a non-isolated singular point, and trajectories starting off the equilibrium line are repelled along the direction of $\mathbf{v}^{(2)}$.

Step 5 · Eigenvectors

For $\lambda_1=0$: $$A\mathbf{v}=0 \implies \begin{pmatrix}3&-2\\-6&4\end{pmatrix}\mathbf{v}=0 \implies v_2=\tfrac{3}{2}v_1.$$ Taking $v_1=2$: $\mathbf{v}^{(1)}=(2,3)^\top$, which spans the line of equilibria $y=\tfrac{3}{2}x$. For $\lambda_2=7$: $$(A-7I)\mathbf{v}=0 \implies \begin{pmatrix}-4&-2\\-6&-3\end{pmatrix}\mathbf{v}=0 \implies v_2=-2v_1.$$ Taking $v_1=1$: $\mathbf{v}^{(2)}=(1,-2)^\top$. Trajectories not lying on the equilibrium line are repelled from it along this direction (lines of slope $-2$).

Step 6 · General solution

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1\begin{pmatrix}2\\3\end{pmatrix} +C_2 e^{7t}\begin{pmatrix}1\\-2\end{pmatrix}.$$ In components: $$x(t)=2C_1+C_2 e^{7t},\qquad y(t)=3C_1-2C_2 e^{7t}.$$

Step 7 · Family of integral curves

Eliminating $e^{7t}$ from $x-2C_1=C_2e^{7t}$ and $\tfrac{3C_1-y}{2}=C_2e^{7t}$ gives $2x+y=7C_1=\text{const}$. The family of integral curves is therefore $$2x+y=K,\quad K\in\mathbb{R},$$ a family of parallel straight lines with slope $-2$, each intersecting the equilibrium line $y=\tfrac{3}{2}x$ at the point $x=\tfrac{2K}{7}$, $y=\tfrac{3K}{7}$.

Step 8 · Behaviour of trajectories

If $C_2=0$ the trajectory is the constant solution $(2C_1,3C_1)$, a point on the equilibrium line. If $C_2\neq0$, since $\lambda_2=7>0$ the term $C_2e^{7t}$ grows without bound as $t\to+\infty$, so the trajectory diverges to infinity along the direction $\mathbf{v}^{(2)}$; as $t\to-\infty$ the trajectory approaches the equilibrium point $(2C_1,3C_1)$. Every non-equilibrium trajectory therefore starts (as $t\to-\infty$) at some point on the equilibrium line and escapes to infinity (as $t\to+\infty$) along a line of slope $-2$.

Step 9 · Conclusion

$$\boxed{ \text{The singular set is the line }y=\tfrac{3}{2}x;\text{ every point on it is a \textbf{non-isolated} (degenerate) singular point.} }$$ The eigenvalues are $\lambda_1=0$ and $\lambda_2=7$. The zero eigenvalue corresponds to the direction of the equilibrium line ($\mathbf{v}^{(1)}=(2,3)^{\top}$), while the positive eigenvalue $\lambda_2=7$ drives all non-equilibrium trajectories away from the equilibrium line along lines of slope $-2$ ($\mathbf{v}^{(2)}=(1,-2)^{\top}$). The family of integral curves is $2x+y=K$, $K\in\mathbb{R}$.

Summary Table

Additional Notes

• Because the coefficient matrix is singular, the equilibria form a line rather than an isolated point; the zero eigenvalue corresponds to the direction of that line and the non-zero eigenvalue governs the transverse behaviour.