### Author Topic: Home Assignment 1  (Read 1769 times)

#### Brittany Palandra ##### Home Assignment 1
« on: January 14, 2019, 10:04:54 PM »
Problem 4 was to find the general solution to $uu_{xy} = u_xu_y$. I used the hint and got $\frac{\partial}{\partial y} \ln u_x = \frac{\partial}{\partial y} \ln u$, but I'm not sure where to go from here.
« Last Edit: January 15, 2019, 04:52:06 PM by Victor Ivrii »

#### Wanying Zhang ##### Re: Home Assignment 1
« Reply #1 on: January 15, 2019, 12:02:54 AM »
I have a guess but also not quite sure if I'm in the right way. Integrate the equation you got
$$\int_ {}{}\frac{(uu_x)_y}{uu_x} dy = \int {}{}\frac{u_y}{u}dy \qquad\qquad\qquad\qquad \color{red}{\text{Error in the left part; the rest incorrect. V.I.}}$$
$$\ln uu_x = \ln u + f(x)$$
where $f$ is any functions that is only of $x$,
$$uu_x = u \cdot g(x)$$
where $g$ is a function of $f$ mentioned above. Divide $u$ from both sides,
$$u_x = g(x)$$
then integrate both sides,
$$u = G(x) + h(y)$$
where $G'(x) = g(x)$, and $h$ is any function that is only of y.
« Last Edit: January 16, 2019, 07:35:04 AM by Victor Ivrii »

#### Wanying Zhang ##### Re: Home Assignment 1
« Reply #2 on: January 16, 2019, 01:42:48 PM »
This is my new try: Follow the hint and I got,
$$\frac{u_{xy}}{u_x} = \frac{u_y}{u}$$
integrate both sides,
$$\ln (u_x) = \ln (u) + f(x)$$
where $f(x)$ is any function only of x. Then, we have,
$$u_x = g(x) \cdot u$$
where $g(x)$ is a function of $f(x)$ (i.e. $g(x) = e^{f(x)}$). Rewrite this equation,
$$\frac{\partial u}{\partial x} = g(x) \cdot u$$
Above is ok, but below is not: you should wright $dx$ and $du$ (because standalone $\partial u$ and $\partial x$ do not make sense), and you need to integrate rather than differentiate
$$\frac{\partial u}{u} = g(x) \partial x$$
$$-\frac{1}{u^2} = g'(x) + h(y)$$
where $h$ is any function that only of y. Therefore,
$$u = \frac{1}{\sqrt{-g'(x) - h(y)}}$$
« Last Edit: January 16, 2019, 03:23:58 PM by Victor Ivrii »

#### JUNJING FAN

• Jr. Member
•  • Posts: 13
• Karma: 0 ##### Re: Home Assignment 1
« Reply #3 on: January 16, 2019, 04:36:55 PM »
Not sure if this will work

#### Victor Ivrii ##### Re: Home Assignment 1
« Reply #4 on: January 16, 2019, 04:57:59 PM »
Junjing, you are right but I am not sure if anyone but me would be able to read your solution

#### JUNJING FAN

• Jr. Member
•  • Posts: 13
• Karma: 0 ##### Re: Home Assignment 1
« Reply #5 on: January 16, 2019, 05:07:33 PM »
Junjing, you are right but I am not sure if anyone but me would be able to read your solution
Okay. Next time I will make it 'skinnier'. Sorry.