16

**Chapter 3 / Re: Chapter 3.1 Theorem 4**

« **on:**February 27, 2022, 07:52:52 PM »

Indeed, to be corrected. Thanks

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

16

Indeed, to be corrected. Thanks

17

Quote

But how does this qualify us to replace the indefinite integral with the definite one?

Did you take Calculus I? Then you

18

We integrate from $t=0$ because for $t=0$ initial conditions are done. $-1<t $ is a domain where $f(x,t)$ is defined

19

Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.

20

It is the same answer for $x>ct$ and $x-ct$. Explain why

21

Indeed, fixed. For consistency added index $_n$ to similar places of Example 4.2.7

Please post in the appropriate subforum

Please post in the appropriate subforum

24

Thanks! Fixed online TB

25

You almost there. Think!

26

- 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

28

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.

29

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

30

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