### Recent Posts

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91
##### 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"
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##### 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 $C=?$
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##### 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 $xe^{-t}$?
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##### 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.

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##### 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
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##### 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 $m(x)$"

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##### 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
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##### 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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##### Chapter 1 / Re: Classification of PDEs
« Last post by Victor Ivrii on January 14, 2022, 01:47:15 PM »
Yes, all linear are also semilinear and all semilinear are also quasilinear. For full mark you need to provide the most precise classification. So, if equation is linear you say "linear", if it is semilinear but not  linear you say "semilinear but not  linear" and so on,... "quasilinear but not  semilinear" and "non-linear and not quasilinear".
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##### Chapter 1 / Re: Classification of PDEs
« Last post by Weihan Luo on January 14, 2022, 11:28:53 AM »