### Author Topic: Deriving equation 7 of section 2.1  (Read 1801 times)

#### Shaghayegh A

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##### Deriving equation 7 of section 2.1
« on: September 24, 2016, 03:20:01 PM »
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
« Last Edit: September 25, 2016, 02:17:04 AM by Victor Ivrii »

#### Victor Ivrii

In red are my corrections, in orange your errrors. Please don't confuse $\partial$ and $d$.