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**Test 1 / Re: Test 1 Wave equation and Heat equation questions **

« **on:**March 01, 2022, 11:21:39 AM »

Yes, you need to perform calculations

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Yes, you need to perform calculations

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Indeed, to be corrected. Thanks

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Quote

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

Did you take Calculus I? Then you

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

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

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

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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

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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)