# The sign of terms in a characteristic equation of a PDE

This was something I noticed as I was trying to practice solving PDEs using the method of characteristics.

The text has the following example: $$\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = 0$$

This should be easy enough. I let $p(x,y) = x$ and solve for $\frac{\partial y}{\partial x}$. When I do that I get: $$\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} =0 \Rightarrow \frac{\partial u}{\partial x} =- x \frac{\partial u}{\partial y} \Rightarrow \frac{\partial u}{\partial x}\frac{\partial y}{\partial u} = -x \frac{\partial u}{\partial y}\frac{\partial y}{\partial u} \Rightarrow \frac{\partial y}{\partial x} = -x$$

This should be easy enough from there, I would have $\int \partial y= - x \partial x \Rightarrow y=-\frac{1}{2}x^2+ C$ and $y+\frac{1}{2}x^2 = C$ so $u(x,y) = y+\frac{1}{2}x^2$. General solution is $f(y+\frac{1}{2}x^2)$ where f is an arbitrary function.

SO what's the problem? The text says that $\frac{\partial y}{\partial x}=x$ not $-x$. Anyone know what's up? I have consistently found that the my HW solutions and text say this should be a different sign than what i get when I just solve for $\frac{\partial y}{\partial x}$. What am I missing? I figure its probably something simple enough.

Thanks.

That is, u + v is also a solution, as is $\alpha u$ for constant $\alpha$.