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**Chapter 3 / Re: Chapter 3.2 Problem 9**

« **on:**February 06, 2022, 02:21:30 PM »

Indeed. You got it!

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Thanks! Fixed online TB

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You almost there. Think!

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- If you do not know this integral you need to refresh Calcuus I. one of basic integrals. Or have a table of basic integrals handy.
- Since $x^2+y^2=c^2$ is a circle, you can substitute $x=c\cos(s)$ and $y=c\sin(s)$ and then observe that $s=D-s$. It gives you the answer, less nicely looking than the one you wrote.
- Expressing $x, y$ through $t,c,d$ you can express $C=c\cos(d)$ and $D=c\sin(d)$ through $x,y,t$ which would give you that nice answer.

Write \cos , \sin , \log .... to produce proper (upright) expressions with proper spacing

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As $x>0$ it is a correct calculation. However $f(xe^{-t})$ re,mains valid for $x<0$ while $f(t-\ln (x))$ does not.

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Yes, there are many answers which are correct because they include arbitrary functions (and in ODE arbitrary constants)

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For the middle region in red, since $\psi(x - \frac{t}{3})$ is undefined,It is defined, because on the line $\{x=-t, t>0\}$ we have not 1 but

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You should ask everybody, not just me. And the question was a bit different anyway: to classify an equation

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Yes, it is correct: $u=H(x)+m(y,z)$ where $H$ and $m$ are arbitrary functions.

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It was a misprint; it should be $e^{u(x',y)}$ in the denominator

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Both signs. What are curves $x^2-y^2=C$?

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Please post non-mathematical questions in Quercus discussions.

Please read description of Quiz 1.

Please read description of Quiz 1.

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Then your best shot would be *continuity condition*.

- either to ask during Prof Kennedy's Office hours
- or to formulate the problem and define everything (that is $u, v, S, \tilde{S}$) here by yourself without screenshots

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It would be really helpfull if you explained where you took this from (if online TextBook--then section and equation number, if lecture then which lecture and which part).