Author Topic: Homework 2: Problem 3  (Read 1554 times)

Jaisen Kuhle

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Homework 2: Problem 3
« on: January 23, 2018, 05:39:23 AM »
For (14), we are given:

$$(x^2+1)yU_{x}+(y^2+1)xU_{y}=0$$

If I didn't make a mistake, the characteristic equation is:

$$ C=\frac{1+x^2}{1+y^2}$$

But now I'm a little confused how to solve:

$$ du=\frac{dx}{y(1+x^2)}$$ Wrong. And be consistent: either $u$ or $U$  V.I.

Anyone have an idea? If we solve y in terms of x we get two values because of the root, so how should we proceed?
« Last Edit: January 23, 2018, 06:05:10 AM by Victor Ivrii »

Jingxuan Zhang

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Re: Homework 2: Problem 3
« Reply #1 on: January 23, 2018, 07:20:20 PM »
Keisen you missed the square root. That cannot be cancelled lightly. Continuity I think is automatic since the characteristic is continuous.

Jaisen Kuhle

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Re: Homework 2: Problem 3
« Reply #2 on: January 23, 2018, 08:00:25 PM »
Keisen you missed the square root. That cannot be cancelled lightly. Continuity I think is automatic since the characteristic is continuous.

Which square root?

Jaisen Kuhle

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Re: Homework 2: Problem 3
« Reply #3 on: January 23, 2018, 08:02:57 PM »
Also, I see a mistake (assuming my characteristic is ok), since it is $\frac{dU}{0}$, we obtain for (14):

$$ U(x,y)=\phi(\frac{1+x^2}{1+y^2})$$

Jingxuan Zhang

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Re: Homework 2: Problem 3
« Reply #4 on: January 23, 2018, 09:50:05 PM »
When you solve the integral curve the one half cannot be cancelled lightly
$$\int\frac{x}{x^{2}+1} \,dx = \ln \sqrt{x^2+1}+C$$
or otherwise what you get is distorted in scale.

Ioana Nedelcu

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Re: Homework 2: Problem 3
« Reply #5 on: January 24, 2018, 12:38:34 AM »
Yeah I got the same solution as Jaisen: $ u = f(\frac{(x^2+1)}{(y^2+1)}) $

For the one with the negative sign, $ u = f((x^2+1)(y^2+1)) $

Jingxuan Zhang

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Re: Homework 2: Problem 3
« Reply #6 on: January 24, 2018, 06:09:21 AM »
But plug in the operation you would have
$$\frac{2xy(x^{2}+1)}{y^{2}+1}\varphi'(\frac{x^{2}+1}{y^{2}+1})+x(x^{2}+1)(y^{2}+1)(\tan^{-1}y)\varphi'(\frac{x^{2}+1}{y^{2}+1})$$

Whereas if $u=\varphi\left(\sqrt{\frac{x^{2}+1}{y^{2}+1}} \right) $ then plugging in you will happily see
$$xy(x^{2}+1)^{1/2}(y^{2}+1)^{-1/2}\varphi'-xy(x^{2}+1)^{1/2}(y^{2}+1)^{-1/2}\varphi'=0.$$

Similarly the other problem you would have product of square root. As I mentioned, when you finish integrating the characteristic you cannot cancel that a half out from both side. That will rescale the constant term and is significant after you take exponential.
« Last Edit: January 24, 2018, 06:26:57 AM by Jingxuan Zhang »

Jaisen Kuhle

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Re: Homework 2: Problem 3
« Reply #7 on: January 24, 2018, 01:01:14 PM »
When you solve the integral curve the one half cannot be cancelled lightly
$$\int\frac{x}{x^{2}+1} \,dx = \ln \sqrt{x^2+1}+C$$
or otherwise what you get is distorted in scale.

I'm not sure how you are solving the integral. Let $$v=1+x^2$$ So $$\frac{dv}{2x}=dx$$ and $$\int\frac{x}{x^{2}+1} \,dx =\int\frac{1}{2v} \,dv = \frac{1}{2}\log|v| + C = \frac{1}{2} \log|1+x^2|+C= \frac{1}{2} \log(1+x^2)+C$$

Victor Ivrii

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Re: Homework 2: Problem 3
« Reply #8 on: January 24, 2018, 01:27:35 PM »
The correct solution was presented by different posters. Still it would be better to start from
$$
\frac{dx}{?} =\frac{dy}{?}=\frac{du}{?}.
$$

Also, are $\phi(z)$ and $\psi (\sqrt{z})$ different where $\phi$ and $\psi$ arbitrary functions and $z>0$?