Examine the singular points of the equation y'=(4y-2x)/(x+y)

Examine the singular points of the equation $$y'=\frac{4y-2x}{x+y}.$$

Step 1 · Symmetric form

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

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 4y-2x=0,\\ x+y=0. \end{cases}$$ From the second equation $x=-y$. Substituting into the first: $$4y-2(-y)=4y+2y=6y=0\implies y=0,\quad x=0.$$ 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 $$\frac{dx}{dt}=P(x,y)=4y-2x,\qquad \frac{dy}{dt}=Q(x,y)=x+y,$$ with coefficient matrix $$A=\begin{pmatrix}-2&4\\1&1\end{pmatrix}.$$

Step 4 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (-2-\lambda)(1-\lambda)-4\cdot1=0.$$ Expanding: $$-2+2\lambda-\lambda+\lambda^2-4=0 \implies \lambda^2+\lambda-6=0 \implies (\lambda+3)(\lambda-2)=0.$$ Thus $$\lambda_1=-3,\qquad \lambda_2=2.$$

Step 5 · Classification of the singular point

The two eigenvalues are real, distinct, and of opposite sign: $$\lambda_1=-3<0<\lambda_2=2.$$ By the standard classification of linear autonomous systems, the origin is a saddle point.

Step 6 · Eigenvectors and separatrices

For $\lambda_1=-3$: $$(A+3I)\mathbf{v}=0 \implies \begin{pmatrix}1&4\\1&4\end{pmatrix}\mathbf{v}=0.$$ Both rows give $v_1+4v_2=0$, so $$\mathbf{v}^{(1)}=\begin{pmatrix}4\\-1\end{pmatrix}.$$ The stable separatrix is tangent at the origin to the line $\ell_1:\; y=-\tfrac{1}{4}x$. Since $\lambda_1<0$, trajectories on this manifold are attracted to the origin as $t\to+\infty$. For $\lambda_2=2$: $$(A-2I)\mathbf{v}=0 \implies \begin{pmatrix}-4&4\\1&-1\end{pmatrix}\mathbf{v}=0.$$ 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 $\ell_2:\; y=x$. Since $\lambda_2>0$, trajectories on this manifold depart from the origin as $t\to+\infty$.

Step 7 · General solution of the linear system

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1 e^{-3t}\begin{pmatrix}4\\-1\end{pmatrix} +C_2 e^{2t}\begin{pmatrix}1\\1\end{pmatrix}.$$ In components: $$x(t)=4C_1 e^{-3t}+C_2 e^{2t},\qquad y(t)=-C_1 e^{-3t}+C_2 e^{2t}.$$

Step 8 · Verification via straight-line solutions

Along $y=-\tfrac{1}{4}x$: substituting into the right-hand side gives $-4$, which does not equal the slope $-\tfrac14$, so the line is only an asymptotic (eigen-)direction of the linearised system, not an exact straight-line solution of the scalar ODE. Along $y=x$: substituting gives $$\frac{4y-2x}{x+y}=\frac{4x-2x}{x+x}=\frac{2x}{2x}=1,$$ which equals the slope of $y=x$, so $y=x$ is indeed an exact straight-line solution of the original scalar ODE, confirming $\ell_2$ as the true unstable separatrix.

Step 9 · Behaviour of integral curves near the origin

The origin is a saddle with a stable manifold tangent to $y=-\tfrac{1}{4}x$ and an unstable manifold that coincides exactly with the line $y=x$. All other trajectories approach the origin along $\ell_1$ as $t\to-\infty$ and escape to infinity along $\ell_2$ as $t\to+\infty$, forming the characteristic hyperbolic pattern of a saddle point.

Step 10 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point; it is a \textbf{saddle point}.} }$$ The eigenvalues $\lambda_1=-3$ and $\lambda_2=2$ are real with opposite signs. The stable separatrix is tangent to the line $y=-\tfrac{1}{4}x$ and the unstable separatrix coincides with the exact straight-line solution $y=x$ of the ODE. 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.