Toronto Math Forum
APM3462015F => APM346Home Assignments => HA1 => Topic started by: Chi Ma 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
$$\frac{u_{xy}}{u_x} = u_y\\
\frac{\partial}{\partial y}\ln(u_x) = u_y\\
u_x = e^ue^{\xi(x)}\\
e^{u}u_x = e^{\xi(x)} \equiv \phi^\prime(x)\\
e^{u} = \phi(x) + \psi(y)\\
u = \ln(\phi(x) + \psi(y))$$

4.
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter1/S1.P.html#problem1.P.4 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter1/S1.P.html#problem1.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}

@Chi
Seems good
@Emily
\begin{equation} ln(u_{x}) = u + f(x)\end{equation}
Then proceed as Chi did

Chi Ma, perfect!