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

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

Step 1 · Symmetric form

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

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} y-2x=0,\\ y=0. \end{cases}$$ From the second equation $y=0$, substituting into the first gives $-2x=0$, hence $x=0$. The unique singular point is $$(x^*,y^*)=(0,0).$$

Step 3 · Associated linear autonomous system

Since $P$ and $Q$ are homogeneous of degree 1, write the equivalent planar system $$\frac{dx}{dt}=P(x,y)=y-2x,\qquad \frac{dy}{dt}=Q(x,y)=y,$$ with coefficient matrix $$A=\begin{pmatrix}-2&1\\0&1\end{pmatrix}.$$

Step 4 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (-2-\lambda)(1-\lambda)-0=0 \implies (\lambda+2)(\lambda-1)=0.$$ Thus $$\lambda_1=-2,\qquad \lambda_2=1.$$

Step 5 · Classification of the singular point

The two eigenvalues are real, distinct, and of opposite sign: $$\lambda_1=-2<0<\lambda_2=1.$$ By the standard classification of linear autonomous systems, the origin is a saddle point.

Step 6 · Eigenvectors and separatrices

For $\lambda_1=-2$: $$(A+2I)\mathbf{v}=0 \implies \begin{pmatrix}0&1\\0&3\end{pmatrix}\mathbf{v}=0.$$ Both rows require $v_2=0$, so $$\mathbf{v}^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}.$$ The corresponding separatrix is tangent to the $x$-axis at the origin. Since $\lambda_1<0$, trajectories on this manifold are attracted to the origin (stable separatrix). For $\lambda_2=1$: $$(A-I)\mathbf{v}=0 \implies \begin{pmatrix}-3&1\\0&0\end{pmatrix}\mathbf{v}=0.$$ The first row gives $-3v_1+v_2=0$, so $$\mathbf{v}^{(2)}=\begin{pmatrix}1\\3\end{pmatrix}.$$ The corresponding separatrix is tangent to the line $y=3x$ at the origin. Since $\lambda_2>0$, trajectories on this manifold depart from the origin (unstable separatrix).

Step 7 · General solution of the linear system

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1 e^{-2t}\begin{pmatrix}1\\0\end{pmatrix} +C_2 e^{t}\begin{pmatrix}1\\3\end{pmatrix}.$$ In components: $$x(t)=C_1 e^{-2t}+C_2 e^{t},\qquad y(t)=3C_2 e^{t}.$$

Step 8 · Behaviour of integral curves near the origin

The saddle point has two invariant straight-line trajectories passing through the origin:

• Stable manifold (along $\mathbf{v}^{(1)}$, the $x$-axis, $y=0$): trajectories approach the origin as $t\to+\infty$ from both sides along the positive and negative $x$-axis.
• Unstable manifold (along $\mathbf{v}^{(2)}$, the line $y=3x$): trajectories depart from the origin as $t\to+\infty$ along $y=3x$.

Step 9 · Verification via the original ODE

Along $y=0$ (and $x\neq 0$) the right-hand side $\tfrac{y-2x}{y}$ is undefined, confirming that $y=0$ is a singular line of the ODE, consistent with the stable separatrix lying along the $x$-axis.

Step 10 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point; it is a \textbf{saddle point}.} }$$ The eigenvalues $\lambda_1=-2$ and $\lambda_2=1$ are real with opposite signs. The stable separatrix is tangent to the $x$-axis and the unstable separatrix is tangent to the line $y=3x$ at the origin. All other trajectories exhibit the characteristic hyperbolic divergence of a saddle.

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.