### Author Topic: HA1-P3  (Read 2361 times)

#### Yumeng Wang

• Full Member
•   • Posts: 20
• Karma: 0 ##### HA1-P3
« on: September 26, 2015, 09:22:18 PM »
This is what I get for question3(d). I used method learned from MAT244 to solve integral linear equation
Sorry for hand writing.

#### Emily Deibert ##### Re: HA1-P3
« Reply #1 on: September 29, 2015, 08:39:58 PM »
3.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.3
a) \begin{equation}
u = \int f(x)dx + g(y)
\end{equation}
b) \begin{equation}
u = (Ce^{2y})x + f(y)
\end{equation}
c) \begin{equation}
u = e^{x+y} + \int f(x)dx + g(y)
\end{equation}
d) I am not very confident in this answer, so hopefully someone is able to explain this question to me! \begin{equation}
u = \frac{e^xe^y}{(1-2y)}
\end{equation}

#### Andrew Lee Chung

• Jr. Member
•  • Posts: 9
• Karma: 0 ##### Re: HA1-P3
« Reply #2 on: September 30, 2015, 04:44:59 PM »
@Emily

d)
\begin{equation}
u_{xy} = 2u_{x}+e^{x+y} \\
Let \ v = u_{x} \\
v_{y} = 2v +e^{x+y}  \\
v-2vy=e^{x+y} +f_{1}(x) \\
v=u_{x}= \frac{e^{x+y}+f_{1}(x)}{1-2y} \\
u= \frac{e^{x+y}+f_{2}(x)}{1-2y} +g(y) \\where \
f_{2}(x) = \int f_{1}(x)dx
\end{equation}

#### Rong Wei

• Sr. Member
•    • Posts: 43
• Karma: 0 ##### Re: HA1-P3
« Reply #3 on: October 01, 2015, 01:45:20 PM »
mengmeng, please use matlab next time

#### Emily Deibert ##### Re: HA1-P3
« Reply #4 on: October 01, 2015, 03:56:35 PM »
Thanks for the help on that one, Andrew. It makes a lot more sense now.

#### Victor Ivrii ##### Re: HA1-P3
« Reply #5 on: October 03, 2015, 05:25:10 AM »
B. $u_{xy}=2u_x\implies u_{xy}/u_x=2$; since l.h.e. is $(\ln (u_x))_y$ we have $\ln u_x = 2y + \ln \phi(x)\implies u_x=\phi(x)y^{2y}\implies u= f(x)e^{2y}+g(x)$.

D.
$u_{xy}=2u_x + e^{x+y}$. Plug $v=u_x$: $v_y = 2v + e^{x+y}$. With respect to $y$ it is a 1st order linear ODE (and $x$ is just some constant here, right?) Partial solution to this ODE is $v=-e^{x+y}$ (you can use formulas but one can just use method of undetermined coefficients as it is a linear ODE with a "special" (quasipolynomial) r.h.e.) and the general $v= -e^{x+y} + f(x) e^{2y}$.

So: $u_x= -e^{x+y} + f(x) e^{2y}$ and integrating by $x$ we get $u=e^{-x+y}+ F(x)e^{2y}+G(y)$ with arbitrary $F,G$

PS Yumeng: you found $v=u_x$ correctly but then erred in the integration. Please, no colour phone snapshots!!! If you cannot type at least scan properly or make pictures properly (black/white with properly positioned camera)
« Last Edit: October 03, 2015, 05:29:52 AM by Victor Ivrii »