### Recent Posts

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91
##### Chapter 2 / Re: Transport Equation Derivation
« Last post by Victor Ivrii on January 19, 2022, 05:06:34 AM »
It would be really helpfull if you explained where you took this from (if online TextBook--then section and equation number, if lecture then which lecture and which part).
92
##### Chapter 2 / Transport Equation Derivation
« Last post by Yifei Hu on January 18, 2022, 07:34:32 PM »
Can anyone help explain where does the Ut term in the second last line come from? Thanks
93
##### Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« Last post by Xiangmin.Z on January 17, 2022, 08:19:53 PM »
$C=xe^{-t}$, so D is a function of $\phi$, therefore $D=\phi({xe^{-t}} )$ ? Yes, but "$D$  is a function of it"
94
##### Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« Last post by Victor Ivrii on January 17, 2022, 07:48:38 PM »
Now it is correct $x=Ce^{t}$ and then $C=?$
95
##### Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« Last post by Xiangmin.Z on January 17, 2022, 05:27:38 PM »
I checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?
96
##### Chapter 2 / Week 2 Lec 1 (Chapter 2) question
« Last post by Xiangmin.Z on January 17, 2022, 04:40:37 PM »
Hello, I have a question about example 4 from W2 L1:
We are given :$u_{t}+xu_{x}=xt$
after calculation we get:
$x=Ce^t$

$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,
but how did we get $D=\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?

Also, why would the initial condition $u|_{t=0} =0$ implies that $\phi({x}) =x$ ?

Thanks.

97
##### Chapter 1 / Re: chapter 1 Problem 4 (1)
« Last post by Victor Ivrii on January 17, 2022, 01:29:44 AM »
Display formulae are surrounded by double dollars and no empty lines. Multiline formulae use special environments (google LaTeX gather align
98
##### Chapter 1 / chapter 1 Problem 4 (1)
« Last post by asdfghj on January 16, 2022, 07:34:18 PM »
$uu_{xy}=u_{x}u_{y}$

$(u_{x}u_{y})/uu_{x}=u_{xy}/u_{x}$

divide both side by$uu_{x}$ and get

$u_{y}/u=u_{xy}/u_{x}$

integrate with respect to y

$\ln{u}+f(x)=\ln{u_{x}}+g(x)$ enough to write one function of $x$

let g(x)-f(x)=n(x)

$u=u_{x}\times n(x)$

$u_{x}/u=n(x)$

$\ln{u}=N(x)+m(y)$

$u=N_{1}(x)\times m(y)$ "another $m(x)$"

99
##### Chapter 1 / Re: home assignment1 Q3(1),(2)
« Last post by Victor Ivrii on January 16, 2022, 05:47:56 PM »
OK. Remarks:

1. Do not use $*$ as a multiplication sign!
2. Do not use LaTeX for italic text (use markdown of the forum--button I)
3. Escape ln, cos, .... : \ln (x) to produce $\ln (x)$ and so on
100
##### Chapter 1 / home assignment1 Q3(1),(2),(3)&(4)
« Last post by asdfghj on January 16, 2022, 04:49:37 PM »
(1):
$u_{xy}=0,denote: v=u_{x}$
$u_{xy}=v_{y}=0$
$v=f(x)$
$u=F(x)+g(y), (let F'(x)=f(x))$

(2):
$u_{xy}=2u_{x}$
let$u_{x} = v$, so
$u_{xy}=v_{y}$
$therefore: v_{y}=v$ integrate on both sides
$v_{y}/v=2$
$2y+f_{1}(x)=\ln(v)$
$v=u_{x}=e^{2y}\times f_2(x)$
let $f_{2}(x)=e^{f_{1}(x)}$
$u=f_{3}(x)\times e^{2y}+g(y)$
where $f'_{3}(x)=f_{2}(x)$

(3):
$u_{xy}=e^{xy}$
$u_{x}=e^{xy}y+f(x)$
$u(x,y)=e^{xy}xy+F(x)+g(y)$

(4)
$u_{xy}=2u_{x}+e^{x+y}$
$u_{xy}=u_{yx}$
$e^{xy}=D(x,y)$
integrate on both sides
$\int{u_{xy}}=\int{2u_{x}+D(x,y)}$
$u_{y}=2u+xD(x+y)+f(y)$
so
$u=u^2+xD(x,y)+F(y)+g(x)$
the general solution is :
$u=u^2+x\times e^{xy}+F(y)+g(x)$

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