Finding the solution form for the PDE $au_x+bu_y+cu=0$ 
Find the solution to the PDE $au_x+bu_y+cu=0$.

Thoughts: Using the substitutions $x^*=ax+by$ and $y^*=bx-ay$, we obtain
$$a\left(\frac{\partial u}{\partial x^*}\frac{\partial x^*}{\partial x}+\frac{\partial u}{\partial y^*}\frac{\partial y^*}{\partial x}\right)+b\left(\frac{\partial u}{\partial x^*}\frac{\partial x^*}{\partial y}+\frac{\partial u}{\partial y^*}\frac{\partial y^*}{\partial y}\right)+cu=0$$
$$a(au_{x^*}+bu_{y^*})+b(bu_{x^*}-au_{y^*})+cu=0$$
$$(a^2+b^2)\frac{\partial}{\partial x^*}u(x^*,y^*)=-cu(x^*,y^*)$$
Now it looks like a separation of variables ODE, but I'm not certain the algebra is analogous, that is, while I can write $dy/dx=f(x)g(y)\Longrightarrow \int g(y)^{-1}dy=\int f(x)dx$, I'm not confident that the same "differential algebra" will work with partial derivatives. I might expand my work to
$$\int \frac1u\,\partial u=-\int\frac c{a^2+b^2}\,\partial x^*$$
$$\ln|u|=-\frac{cx^*}{a^2+b^2}+f_1(y^*)$$
$$u(x,y)=f(y)e^{-cx/(a^2+b^2)}$$
for some function $f$, but, again, why should I expect this manipulation of differentials to work?
Question: Is my reasoning sound, and does it produce the correct answer?
 A: The answer is correct (modulo a missing change of variables in the end, see the bottom line of this post), and the reasoning is essentially good. I would only advise against the use of symbols like that: 
$$\int\ldots \partial x.$$
Indeed, the symbol "$\partial (\text{something})$" alone does not really make any sense, unlike "$d(\text{something})$" which is interpreted as a total differential. 
I would write as follows. Once you get to the equation $$\frac{\partial u}{\partial x^\star}=Cu, $$
(with $C=-c/ (a^2+b^2)$), fix a value of $y^\star$ and define an auxiliary function $v(x^\star)=u(x^\star, y^\star)$. This "partial function" satisfies the equation 
$$\frac{d v}{d x^\star}(x^\star)=Cv(x^\star), $$
and you may now proceed as you did.
Of course, your result will be expressed in the modified coordinate system $(x^\star, y^\star)$. If you want the result in the original coordinate system $(x, y)$ you will have to undo the change of variables you made in the beginning.
A: Just from inspection, it appears $u(x,y)=Ae^{ax/c}+Be^{by/c}$ is a family of solutions.
