Solving homogeneous differential equation in symmetric form Let $g: \mathbb R \rightarrow \mathbb R$ be a differentiable and integrable function. The integral curve of the differential equation is:
$(y + g(x))dx + (x - g(y))dy = 0$
that passes through the point $(1, 1)$ must also pass through which of the following points?
$(0, 0),$
$(2, 1/2),$
$(1/2, 2),$
$(-1, -1),$ or 
$(0, 1)$
 A: If $f$ is a primitime function for $g$ then the equation is $d(xy+f(x)-f(y))=0$, so the integral curve passing through $(1,1)$ is $xy+f(x)-f(y)=1$. If $f=0$ (i.e. $g=0$) then the solution passes only through $(2,1/2)$ and $(1/2,2)$ ($(-1,-1)$ is not there - we should only take one branch of the hyperbola $xy=1$). On the other hand, if $f(2)\neq f(1/2)$, then $(2,1/2)$ and $(1/2,2)$ are not on the curve. So none of the points must lie on the curve.
A: The integral curve can be rewritten as
$$ d(xy) + g(x)dx - g(y)dy = 0 $$
It is given that the curve passes through (1,1). Let $(x_0,y_0)$ be any other point on the curve. Then, integrating the above equation
$$\int_{1}^{x_0y_0} d(xy) + \int_1^{x_0}g(x)dx-\int_1^{y_0}g(y)dy = C $$
As the variable of integration is immaterial, 
$$x_0y_0 -1 +\int_{y_0}^{x_0}g(t)dt =C $$
As $(x_0,y_0) = (1,1)$ satisfies the equation, we can conclude that $C=0$.
In order for a $(x_0,y_0)$ to satisfy the above equation for all $g$, it is clear that $x_0 = y_0$ so that the integral in evaluates to zero regardless of $g$. \
So, the desired solution must satisfy two conditions
1.$x_0 = y_0$
2.$x_0y_0 -1 = 0$ 
Clearly, $(-1,-1)$ is the only possible answer. 
