### Author Topic: Web bonus problem : Week 3 (#5)  (Read 3338 times)

#### Victor Ivrii

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##### Web bonus problem : Week 3 (#5)
« on: September 26, 2015, 12:53:45 PM »

#### Zaihao Zhou

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##### Re: Web bonus problem : Week 3 (#5)
« Reply #1 on: September 29, 2015, 12:04:08 AM »
Part A

u_{xy} = u_{x}u_{y}u^{-1}

Rearrange the function, we have: $$uu_{xy} = u_{x}u_{y}$$
Divide both sides by $$uu_{x}$$, we have $$\frac{u_{xy}}{u_{x}} = \frac{u_{y}}{u}$$

which is $$(lnu_{x})_{y} = (lnu)_{y}$$.
Integrate and do some arrangements we have $$lnu = \int{e^{f(x)}dx}+g(y)$$
Thus $$u = e^{\int{e^{f(x)}dx}}e^{g(y)}$$

Part B arrange we have $$(lnu_{x})_{y} = u_{y}$$
Integrate and arrange we have $$-e^{-u} = \int{e^{f(x)}dx}+g(y)$$
thus we have $$u = -ln[-(\int{e^{f(x)}dx}+g(y))]$$

Part C
« Last Edit: September 29, 2015, 09:37:12 AM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Web bonus problem : Week 3 (#5)
« Reply #2 on: September 29, 2015, 09:40:49 AM »
Part A -- after you got $\bigl(\ln (u_x)\bigr)_y = \bigl(\ln (u)\bigr)_y$ integration gives $\ln (u_x)= \ln (u) +\ln \phi(x)$ (we look for a simple form) $\implies u_x= \phi(x) u\implies (\ln (u))_x =\phi(x)\implies \ln (u)= f(x)+g(y)$ with $f$ primitive of of $\phi$.

So: $u =F(x)G(y)$ where $F=e^f, G=e^g$ are arbitrary functions. You did correct but you need to jump to simple forms all the time

Part B -- ditto, but again $u=-\ln (F(x)+G(y))$ is a simpler form.

Write \ln (and so on)
« Last Edit: October 03, 2015, 04:34:08 AM by Victor Ivrii »