Examine the singular points of the equation…

Examine the singular points of the equation $$\frac{dy}{dx}=\frac{4x-3y}{x-2y}.$$

Step 1 · Symmetric form

Write the equation as $$P(x,y)\,dx - Q(x,y)\,dy = 0,$$ with $$P(x,y)=4x-3y, \qquad Q(x,y)=x-2y.$$

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 4x-3y=0,\\ x-2y=0. \end{cases}$$ From the second equation $x=2y$. Substituting into the first: $$4(2y)-3y=8y-3y=5y=0 \implies y=0,\quad x=0.$$ The unique singular point is $$(x^*,y^*)=(0,0).$$

Step 3 · Linearisation at the origin

Since $P$ and $Q$ are already linear (homogeneous of degree 1), the equation is exactly of the form studied in the theory of linear autonomous systems. Write the equation as the planar system $$\frac{dx}{dt}=P(x,y)=4x-3y,\qquad \frac{dy}{dt}=Q(x,y)=x-2y,$$ with coefficient matrix $$A=\begin{pmatrix}4&-3\\1&-2\end{pmatrix}.$$

Step 4 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (4-\lambda)(-2-\lambda)-(-3)(1)=0.$$ Expanding: $$(4-\lambda)(-2-\lambda)+3=0 \implies -8-4\lambda+2\lambda+\lambda^2+3=0 \implies \lambda^2-2\lambda-5=0.$$ Using the quadratic formula: $$\lambda=\frac{2\pm\sqrt{4+20}}{2}=\frac{2\pm\sqrt{24}}{2}=1\pm\sqrt{6}.$$ Thus $$\lambda_1=1+\sqrt{6}\approx 3.449,\qquad \lambda_2=1-\sqrt{6}\approx -1.449.$$

Step 5 · Classification of the singular point

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

Step 6 · Eigenvectors and separatrices

For $\lambda_1=1+\sqrt{6}$: $$(A-\lambda_1 I)\mathbf{v}=0 \implies \begin{pmatrix}3-\sqrt{6}&-3\\1&-3-\sqrt{6}\end{pmatrix}\mathbf{v}=0.$$ From the first row: $(3-\sqrt{6})v_1=3v_2$, so we may take $$\mathbf{v}^{(1)}=\begin{pmatrix}3\\3-\sqrt{6}\end{pmatrix}.$$ For $\lambda_2=1-\sqrt{6}$: $$(A-\lambda_2 I)\mathbf{v}=0 \implies \begin{pmatrix}3+\sqrt{6}&-3\\1&-3+\sqrt{6}\end{pmatrix}\mathbf{v}=0.$$ From the first row: $(3+\sqrt{6})v_1=3v_2$, so we may take $$\mathbf{v}^{(2)}=\begin{pmatrix}3\\3+\sqrt{6}\end{pmatrix}.$$ The two separatrices (stable and unstable manifolds) are tangent at the origin to the lines spanned by $\mathbf{v}^{(1)}$ and $\mathbf{v}^{(2)}$ respectively.

Step 7 · General solution of the linear system

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1 e^{(1+\sqrt{6})\,t}\begin{pmatrix}3\\3-\sqrt{6}\end{pmatrix} +C_2 e^{(1-\sqrt{6})\,t}\begin{pmatrix}3\\3+\sqrt{6}\end{pmatrix}.$$ Integral curves of the original ODE are obtained by eliminating $t$.

Step 8 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point; it is a \textbf{saddle point}.} }$$ The eigenvalues $\lambda_{1,2}=1\pm\sqrt{6}$ are real with opposite signs, confirming the saddle character. Two separatrices pass through the origin: along $\mathbf{v}^{(1)}$ trajectories move away from the origin ($\lambda_1>0$), while along $\mathbf{v}^{(2)}$ trajectories approach it ($\lambda_2<0$). All other trajectories approach the origin along the unstable separatrix direction and then recede to infinity.

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.