Examine the singular points of the equation…

Examine the singular points of the equation $$\frac{dy}{dx}=\frac{x+y}{2x}.$$

Step 1 · Rewriting as a homogeneous-type system

Write the equation in the symmetric form $$P(x,y)\,dx - Q(x,y)\,dy = 0,$$ where we set $$(x+y)\,dx - 2x\,dy = 0,$$ so $P(x,y)=x+y$ and $Q(x,y)=2x$.

Step 2 · Locating the singular point

A singular (critical) point is a point $(x^*,y^*)$ at which both $P$ and $Q$ vanish simultaneously: $$P(x^*,y^*)=0 \quad\text{and}\quad Q(x^*,y^*)=0.$$ From $Q=2x=0$ we get $x^*=0$, and then $P=x^*+y^*=y^*=0$. Hence the unique singular point is $$(x^*,y^*)=(0,0).$$

Step 3 · Substitution $y = vx$ (homogeneous reduction)

For $x\neq0$ introduce $y=vx$, so $dy=v\,dx+x\,dv$. Substituting: $$\frac{dy}{dx}=\frac{x+y}{2x}=\frac{x+vx}{2x}=\frac{1+v}{2}.$$ Since $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$, we obtain $$v + x\frac{dv}{dx} = \frac{1+v}{2}.$$ Rearranging: $$x\frac{dv}{dx} = \frac{1+v}{2} - v = \frac{1+v-2v}{2} = \frac{1-v}{2}.$$

Step 4 · Separating variables and integrating

Separating variables (for $v\neq1$): $$\frac{dv}{1-v}=\frac{dx}{2x}.$$ Integrating both sides: $$-\ln|1-v|=\frac{1}{2}\ln|x|+C_1 \implies \ln|1-v|=-\frac{1}{2}\ln|x|+C_2.$$ Exponentiating: $$|1-v|=\frac{A}{|x|^{1/2}}, \quad A>0,$$ or, absorbing the sign into the constant $A\in\mathbb{R}\setminus\{0\}$: $$1-v=\frac{A}{x^{1/2}}.$$

Step 5 · General solution in terms of $x$ and $y$

Substituting back $v=y/x$: $$1-\frac{y}{x}=\frac{A}{\sqrt{x}} \implies \frac{x-y}{x}=\frac{A}{\sqrt{x}} \implies x-y=A\sqrt{x}.$$ Hence the general solution is $$\boxed{x - y = A\sqrt{x}}, \quad A\in\mathbb{R},$$ or equivalently $y = x - A\sqrt{x}$.

Step 6 · Behaviour of integral curves near the singular point

All integral curves have the form $y=x-A\sqrt{x}$. As $x\to0^+$: $$y \to 0 \quad \text{for every value of } A,$$ so every trajectory passes through (or approaches) the origin. The slope of each curve at the origin can be analysed by writing $$\frac{y}{x}=1-\frac{A}{\sqrt{x}}\to\pm\infty \quad (A\neq0),$$ meaning that the curves become tangent to the $y$-axis as $x\to0^+$. The singular point $(0,0)$ is therefore an irregular (non-elementary) singular point: infinitely many integral curves arrive at the origin, all tangent to the vertical line $x=0$.

Step 7 · Conclusion

$$\boxed{ (x^*, y^*) = (0,0) \text{ is the unique singular point of the equation.} }$$ The general solution $y=x-A\sqrt{x}$ shows that every integral curve passes through the origin, which is an irregular singular point (a node-like singularity at which the standard uniqueness theorem fails because the right-hand side $\tfrac{x+y}{2x}$ is discontinuous at $x=0$).

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.