Author Topic: Ut+xUx=0  (Read 3924 times)

Yifei Hu

  • Full Member
  • ***
  • Posts: 15
  • Karma: 0
    • View Profile
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

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
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.