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

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

Step 1 · Symmetric form

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

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} y=0,\\ x=0. \end{cases}$$ 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 $$\frac{dx}{dt}=P(x,y)=y,\qquad \frac{dy}{dt}=Q(x,y)=x,$$ with coefficient matrix $$A=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$

Step 4 · Eigenvalues of $A$

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

Step 5 · Classification of the singular point

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

Step 6 · Eigenvectors and separatrices

For $\lambda_1=-1$: $$(A+I)\mathbf{v}=0 \implies \begin{pmatrix}1&1\\1&1\end{pmatrix}\mathbf{v}=0.$$ Both rows give $v_1+v_2=0$, so $$\mathbf{v}^{(1)}=\begin{pmatrix}1\\-1\end{pmatrix}.$$ The stable separatrix is tangent at the origin to the line $\ell_1:\; y=-x$. For $\lambda_2=1$: $$(A-I)\mathbf{v}=0 \implies \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf{v}=0.$$ Both rows give $v_1=v_2$, so $$\mathbf{v}^{(2)}=\begin{pmatrix}1\\1\end{pmatrix}.$$ The unstable separatrix is tangent at the origin to the line $\ell_2:\; y=x$.

Step 7 · Exact solution of the scalar ODE

The original equation $y'=y/x$ is separable. Separating variables: $$\frac{dy}{y}=\frac{dx}{x} \implies \ln|y|=\ln|x|+C \implies y=Kx,\quad K\in\mathbb{R}\setminus\{0\}.$$ The general solution consists of all straight lines through the origin: $y=Kx$, $K\in\mathbb{R}$. Every such line passes through the singular point $(0,0)$.

Step 8 · Verification of separatrices

Setting $K=-1$ gives $\ell_1:\,y=-x$, with slope $y/x=-1$: \checkmark. Setting $K=1$ gives $\ell_2:\,y=x$, with slope $y/x=1$: \checkmark. Both separatrices are exact straight-line solutions of the scalar ODE, consistent with the system analysis.

Step 9 · Nature of the singular point — degenerate case

While the eigenvalue analysis of the associated system classifies the origin as a saddle, the scalar ODE $y'=y/x$ presents a degenerate situation: its general solution $y=Kx$ is a one-parameter family of straight lines all passing through the origin. Every line through the origin is an integral curve, and uniqueness of solutions fails there because $y/x$ is homogeneous of degree zero. The singular point $(0,0)$ is an improper node (also called a stellar or isotropic node) of the scalar ODE, despite the saddle classification of the symmetrised system.

Step 10 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point.} }$$ The general solution $y=Kx$ is a family of lines through the origin. Every integral curve passes through the singular point, and uniqueness of the initial-value problem fails there. The eigenvalues of the associated linear system are $\lambda_{1,2}=\pm1$ (real, opposite sign), giving a saddle for the system; however, for the scalar ODE the origin behaves as an improper (stellar) node through which every direction is a solution trajectory.

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.