Author Topic: home assignment1 Q3(1),(2),(3)&(4)  (Read 1302 times)

asdfghj

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
home assignment1 Q3(1),(2),(3)&(4)
« 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 :
$u=u^2+x\times e^{xy}+F(y)+g(x)$





« Last Edit: January 16, 2022, 07:15:52 PM by zzeng »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: home assignment1 Q3(1),(2)
« Reply #1 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