### Author Topic: problem 5 (23)  (Read 1405 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

• Jr. Member
•  • Posts: 13
• Karma: 0 ##### Re: problem 5 (23)
« 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

• Jr. Member
•  • Posts: 13
• Karma: 0 ##### Re: problem 5 (23)
« 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

• Jr. Member
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• Karma: 0 ##### Re: problem 5 (23)
« 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.