Prove that if the singular point of the equation (ax+by)…

Prove that if the singular point of the equation $$(ax+by)\,dx+(mx+ny)\,dy=0$$ is a centre, then the equation is an exact differential. Show that the converse is not true.

Step 1 · Setup

Write $P=ax+by$, $Q=mx+ny$. Exactness requires $\partial P/\partial y=\partial Q/\partial x$, i.e.\ $b=m$.

Step 2 · Correct correspondence between the ODE and its system

The equation $y'=-P/Q$ has the same orbits as the system $$\dot{x}=Q=mx+ny,\qquad\dot{y}=-P=-(ax+by),$$ with coefficient matrix $$B=\begin{pmatrix}m&n\\-a&-b\end{pmatrix}.$$

Step 3 · Characteristic equation of the associated system

$$\det(B-\lambda I)=\lambda^2+(b-m)\lambda+(an-mb)=0.$$

Step 4 · Conditions for $(0,0)$ to be a centre

A centre requires purely imaginary eigenvalues: $$\operatorname{tr}(B)=m-b=0\implies m=b,$$ $$\det(B)=an-mb=an-m^2>0.$$ Hence the centre condition is $m=b$ and $an-m^2>0$.

Step 5 · Centre condition implies exactness

The condition $m=b$ is exactly the exactness condition $\partial P/\partial y=b=m=\partial Q/\partial x$. Hence a centre forces the equation $(ax+by)\,dx+(mx+ny)\,dy=0$ to be an exact differential.

Step 6 · Explicit first integral

Since $b=m$, integrating gives the first integral $$F(x,y)=\frac{a}{2}x^2+bxy+\frac{n}{2}y^2=K,\quad K>0.$$ This quadratic form is positive definite (closed elliptical level curves) exactly when $a>0$ and $an-m^2>0$, matching the second centre condition.

Step 7 · The converse is false — counterexample

Exactness only requires $b=m$, with no restriction on $an-m^2$. Take $a=1$, $b=m=0$, $n=-1$: the equation $x\,dx-y\,dy=0$ is exact ($\partial P/\partial y=0=\partial Q/\partial x$), with first integral $x^2-y^2=C$ — hyperbolas, so the origin is a saddle point, not a centre. The associated matrix $B=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$ has eigenvalues $\lambda=\pm1$ (real, opposite sign), confirming the saddle.

Step 8 · Conclusion

$$\boxed{ \text{Centre }\implies\text{ exact equation}\quad (m=b\text{ is necessary for a centre and sufficient for exactness).} }$$ $$\boxed{ \text{Exact equation }\not\Rightarrow\text{ centre.} }$$ The singular point of $(ax+by)\,dx+(mx+ny)\,dy=0$ is a centre if and only if $m=b$ and $an-m^2>0$. The condition $m=b$ is precisely the exactness condition, so every centre forces exactness. The converse fails because exactness ($m=b$) does not guarantee $an-m^2>0$: if $an-m^2\leq0$ the origin is a node or saddle rather than a centre, as shown by the counterexample $x\,dx-y\,dy=0$.

Summary Table

Additional Notes

• The argument hinges on relating the scalar equation to its associated planar system and translating the geometric property (centre, exactness, symmetry) into an algebraic condition on the trace and determinant of the corresponding matrix.