 Author Topic: problem 5 (23)  (Read 2345 times)

Wanying Zhang problem 5 (23)
« on: January 20, 2019, 12:53:01 AM »
I have trouble solving this problem: $yu_x - xu_y = x^2$. I know the characteristic equation is $\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}$ and then have $C = \frac{x^2}{2} + \frac{y^2}{2}$. Then the following should be the integration relative to $du$, but either $\frac{du}{dx}$ or $\frac{du}{dy}$ will contain not only one variable, like $\frac{du}{dx} = \frac{x^2}{y}$ contain both $x$ and $y$. I wonder if $x$ and $y$ are independent here. If not, should I rewrite the expression $C = \frac{x^2}{2} + \frac{y^2}{2}$ in order to get the expression of y in terms of x , and then applies it into the integration relative to $du$? Any reply would be appreciated.

Victor Ivrii Re: problem 5 (23)
« Reply #1 on: January 20, 2019, 04:47:13 AM »
Sure, $x$ and $y$ are not independent along integral curves. To proceed you need to parametrize the integral curve. Think: what is the best way to parametrize it?

Wanying Zhang Re: problem 5 (23)
« Reply #2 on: January 20, 2019, 11:53:29 AM »
I tried to use the trigonometric identities to solve this problem and think I got the correct answer, which is $-\frac{1}{2} xy + (\frac{1}{2})(x^2 + y^2)arcsin(\frac{y}{\sqrt{x^2 + y^2}})$. But then I'm confused about "In one instance solution does not exist" at end of this problem because I obtain solutions for all 4 subproblems, except two of them with arbitrary function. I wonder which subproblem may not have solutions.

Victor Ivrii Re: problem 5 (23)
« Reply #3 on: January 20, 2019, 12:32:48 PM »
I obtain solutions for all 4 subproblems, except two of them with arbitrary function.
No, your solution is incorrect because it is not a continuous single-valued function. I gave you a hint: what is the natural parameter along integral curves?

JUNJING FAN

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« Reply #4 on: January 27, 2019, 12:11:13 PM »
Heller professor
By parametrizing Y in terms of X, do we need to put a plus/minus sign in front of the root? If yes, does that mean when we put down the final solution, we need to include plus and minus as well?

Victor Ivrii Re: problem 5 (23)
« Reply #5 on: January 27, 2019, 07:07:51 PM »
There is NO root. You need to parametrize before integration

JUNJING FAN

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« Reply #6 on: January 27, 2019, 07:47:30 PM »
There is NO root. You need to parametrize before integration

but, when we parametrize Y in terms of X, don't we have to use $$y^2+x^2=C$$
and thus $$y= +/- \sqrt{C-x^2}$$?

MikeMorris

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« Reply #7 on: January 27, 2019, 08:43:49 PM »
By parameterizing, the professor means to express x and y in terms of some other parameter. Since the integral curves are circles, think about what is normally used to parameterize a circle in a very easy way - it's essentially a change of variable.