Examine the singular points of the system dotx=x, doty=2x-y

Examine the singular points of the system $$\dot{x}=x,\qquad \dot{y}=2x-y.$$

Step 1 · Locating the singular point

Set both right-hand sides equal to zero: $$\begin{cases} x=0,\\ 2x-y=0. \end{cases}$$ From the first equation $x=0$, substituting into the second gives $y=0$. The unique singular point is $$(x^*,y^*)=(0,0).$$

Step 2 · Coefficient matrix

The system is linear with coefficient matrix $$A=\begin{pmatrix}1&0\\2&-1\end{pmatrix}.$$

Step 3 · Eigenvalues of $A$

The matrix $A$ is lower triangular, so the eigenvalues are the diagonal entries: $$\lambda_1=1,\qquad \lambda_2=-1.$$

Step 4 · Classification of the singular point

The two eigenvalues are real, distinct, and of opposite sign: $$\lambda_2=-1<0<\lambda_1=1.$$ The origin is therefore a saddle point.

Step 5 · Eigenvectors and separatrices

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

Step 6 · General solution

$$\begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1 e^{t}\begin{pmatrix}1\\1\end{pmatrix} +C_2 e^{-t}\begin{pmatrix}0\\1\end{pmatrix}.$$ In components: $$x(t)=C_1 e^{t},\qquad y(t)=C_1 e^{t}+C_2 e^{-t}.$$

Step 7 · Family of integral curves

Eliminating $t$: since $x=C_1 e^t$, we have $e^t=x/C_1$ and $e^{-t}=C_1/x$ (for $C_1\neq0$). Substituting into $y$: $$y=x+\frac{C_1C_2}{x}.$$ Setting $K=C_1C_2$, the family of integral curves is $$xy-x^2=K,\quad K\in\mathbb{R},$$ or equivalently $x(y-x)=K$. For $K=0$: either $x=0$ (stable separatrix) or $y=x$ (unstable separatrix), consistent with the eigenvector analysis.

Step 8 · Behaviour of trajectories near the origin

The saddle point has two invariant trajectories through the origin: the unstable manifold $y=x$ (departing as $t\to+\infty$) and the stable manifold $x=0$ (approaching as $t\to+\infty$). All other trajectories approach the origin along $x=0$ as $t\to-\infty$ and escape to infinity along $y=x$ as $t\to+\infty$, tracing the hyperbolic curves $x(y-x)=K$.

Step 9 · Conclusion

$$\boxed{ (x^*,y^*)=(0,0)\text{ is the unique singular point; it is a \textbf{saddle point}.} }$$ The eigenvalues are $\lambda_{1,2}=\pm1$ (real, opposite sign). The unstable separatrix lies along the line $y=x$ and the stable separatrix lies along the $y$-axis ($x=0$). The family of integral curves is $x(y-x)=K$, $K\in\mathbb{R}$.

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.