In the section variable coefficients of section 2.1, we have
$$
au_t+bu_x=f\tag{6}
$$
Then we have
$$
\frac{\partial u}{\partial t}dt+ \color{orange}{\frac{\partial x}{\partial t}}dt \frac{\partial u}{\partial x}=u
\tag{*}$$
No, $\frac{d x}{d t}$
I assume the $dt$ cancels with the $\partial t$ in the $\frac{\partial x}{\partial t}dt \frac{\partial u}{\partial x} $ part because the textbook says we get
$$u_t dt+dx u_x =du$$
Wrong conclusion due to your error in (*)
Why doesn't the $dt$ cancel the $\partial t $ in $\frac{\partial u}{\partial t}dt$ to give us $du+dxu_x =du$?
Calculus II
Also, to derive $$\frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}\tag{7}$$ from (6) why don't we just compare (6) to
$\frac{du}{dt}=\frac{\partial u}{\partial t}+\color{orange}{\frac{\partial x}{\partial t}}\frac{\partial u}{\partial x}$ (chain rule) and conclude that $\frac{\partial t}{a}=\frac{\partial x}{b}$ and $\frac{dt}{a}=\frac{du}{f} \implies \frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}$ (7) instead of doing all that work?
The same mistake; also there should be $\frac{d t}{a}=\frac{d x}{b}$and if corrected it would be exactly what we do
Thanks
