Examine the singular points of the equation y'=(2x+y)/(3x+4y)

Examine the singular points of the equation $$y'=\frac{2x+y}{3x+4y}.$$

Step 1 · Symmetric form

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

Step 2 · Locating the singular point

Set $P=0$ and $Q=0$ simultaneously: $$\begin{cases} 2x+y=0,\\ 3x+4y=0. \end{cases}$$ From the first equation $y=-2x$. Substituting into the second: $$3x+4(-2x)=3x-8x=-5x=0\implies x=0,\quad y=0.$$ 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 $$\frac{dx}{dt}=P(x,y)=2x+y,\qquad \frac{dy}{dt}=Q(x,y)=3x+4y,$$ with coefficient matrix $$A=\begin{pmatrix}2&1\\3&4\end{pmatrix}.$$

Step 4 · Eigenvalues of $A$

The characteristic equation is $$\det(A-\lambda I)=0 \implies (2-\lambda)(4-\lambda)-3=0.$$ Expanding: $$8-2\lambda-4\lambda+\lambda^2-3=0 \implies \lambda^2-6\lambda+5=0 \implies (\lambda-1)(\lambda-5)=0.$$ Thus $$\lambda_1=1,\qquad \lambda_2=5.$$

Step 5 · Classification of the singular point

Both eigenvalues are real, distinct, and positive: $$0<\lambda_1<\lambda_2.$$ By the standard classification of linear autonomous systems, the origin is an unstable node.

Step 6 · Eigenvectors

For $\lambda_1=1$: $$(A-I)\mathbf{v}=0 \implies \begin{pmatrix}1&1\\3&3\end{pmatrix}\mathbf{v}=0.$$ Both rows give $v_1+v_2=0$, so $$\mathbf{v}^{(1)}=\begin{pmatrix}1\\-1\end{pmatrix}.$$ For $\lambda_2=5$: $$(A-5I)\mathbf{v}=0 \implies \begin{pmatrix}-3&1\\3&-1\end{pmatrix}\mathbf{v}=0.$$ Both rows give $-3v_1+v_2=0$, so $$\mathbf{v}^{(2)}=\begin{pmatrix}1\\3\end{pmatrix}.$$

Step 7 · General solution of the linear system

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1 e^{t}\begin{pmatrix}1\\-1\end{pmatrix} +C_2 e^{5t}\begin{pmatrix}1\\3\end{pmatrix}.$$

Step 8 · Behaviour of integral curves

The two eigendirections determine the nodal axes: $$\ell_1:\; y=-x \quad (\text{along }\mathbf{v}^{(1)},\;\lambda_1=1), \qquad \ell_2:\; y=3x \quad (\text{along }\mathbf{v}^{(2)},\;\lambda_2=5).$$ Since both eigenvalues are positive, all trajectories move away from the origin as $t\to+\infty$. As $t\to-\infty$ every trajectory (except those on $\ell_2$) approaches the origin tangentially to the slower eigendirection $\ell_1$ (the line $y=-x$), which is the characteristic direction of an unstable node. Trajectories lying exactly on $\ell_2$ approach the origin along $y=3x$.

Step 9 · Slope of integral curves at the origin

Along $\ell_1$ the slope is $dy/dx=-1$ and along $\ell_2$ the slope is $dy/dx=3$; checking against the right-hand side confirms the nodal structure is consistent with the eigenvalue analysis: $$\left.\frac{2x+y}{3x+4y}\right|_{y=-x}=\frac{x}{-x}=-1.$$

Step 10 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point; it is an \textbf{unstable node}.} }$$ The eigenvalues $\lambda_1=1$ and $\lambda_2=5$ are both real, distinct, and positive. All trajectories depart from the origin, approaching it tangentially to the line $y=-x$ (slower eigendirection) as $t\to-\infty$, while the faster eigendirection corresponds to $y=3x$.

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.