Author Topic: HA1-P4  (Read 671 times)

Chi Ma

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HA1-P4
« on: September 27, 2015, 11:59:03 AM »
This is what I got for Problem 4.

Problem 4a
$$\frac{u_{xy}}{u_x} = \frac{u_y}{u}\\
\frac{\partial}{\partial y}\ln(u_x) = \frac{\partial}{\partial y}\ln(u)\\
u_x = ue^{\xi(x)}\\
\frac{1}{u}\frac{\partial{u}}{\partial x} = e^{\xi(x)} \equiv \phi^\prime(x)\\
\ln(u) = \phi(x) + \psi(y)\\
u = e^{\phi(x)}e^{\psi(y)}$$

Problem 4b
$$\frac{u_{xy}}{u_x} = 2\frac{u_y}{u}\\
\frac{\partial}{\partial y}\ln(u_x) = 2\frac{\partial}{\partial y}\ln(u)\\
u_x = u^2e^{\xi(x)}\\
\frac{1}{u^{2}}\frac{\partial{u}}{\partial x} = e^{\xi(x)} \equiv -\phi^\prime(x)\\
\frac{1}{u} = \phi(x) + \psi(y)\\
u = \frac{1}{\phi(x) + \psi(y)}$$

Problem 4c
$$\frac{u_{xy}}{u_x} = u_y\\
\frac{\partial}{\partial y}\ln(u_x) = u_y\\
u_x = e^ue^{\xi(x)}\\
e^{-u}u_x = e^{\xi(x)} \equiv -\phi^\prime(x)\\
e^{-u} = \phi(x) + \psi(y)\\
u = -\ln(\phi(x) + \psi(y))$$
« Last Edit: September 27, 2015, 12:39:06 PM by Chi Ma »

Emily Deibert

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Re: HA1-P4
« Reply #1 on: September 29, 2015, 08:40:10 PM »
4.
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.P.html#problem-1.P.4
a) \begin{equation}
u = e^{x + g(y)}
\end{equation}
    b) \begin{equation}
u = e^{e^2x + g(y)}
\end{equation}
    c) I am a little confused about this question, so I don't have an answer yet. Below is my process so far: \begin{equation}
u_{xy} = u_{y}u_{x} \Rightarrow
\frac{u_{xy}}{u_{x}} = u_{y} \Rightarrow
ln(u_{x}) = u_{y} \Rightarrow
\end{equation}

Andrew Lee Chung

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Re: HA1-P4
« Reply #2 on: September 30, 2015, 04:53:04 PM »
@Chi
Seems good

@Emily

\begin{equation} ln(u_{x}) = u + f(x)\end{equation}

Then proceed as Chi did

Victor Ivrii

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Re: HA1-P4
« Reply #3 on: October 03, 2015, 05:34:08 AM »
Chi Ma, perfect!