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### Topics - Chi Ma

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##### HA1 / 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
$$\frac{u_{xy}}{u_x} = u_y\\ \frac{\partial}{\partial y}\ln(u_x) = u_y\\ u_x = e^ue^{\xi(x)}\\ e^{-u}u_x = e^{\xi(x)} \equiv -\phi^\prime(x)\\ e^{-u} = \phi(x) + \psi(y)\\ u = -\ln(\phi(x) + \psi(y))$$

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