Author Topic: Ut+xUx=0  (Read 317 times)

Yifei Hu

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Ut+xUx=0
« on: February 01, 2022, 10:08:42 AM »
When solving this problem, I proceed as follow:
$$\frac{dt}{1}=\frac{dx}{x}=\frac{du}{0}$$
Hence, U does not depend on x and t, integrate on first part of equation:
$$t=ln(x)+C$$
I did not take exponential on both sides to get $e^t=Cx$ but I directly use $C=t-ln(x)$ and got $U=f(t-ln(x))$. Can anyone help me identify why this calculation is wrong?

Victor Ivrii

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Re: Ut+xUx=0
« Reply #1 on: February 01, 2022, 12:16:37 PM »
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.