APM346-2022S > Chapter 2

Ut+xUx=0

(1/1)

**Yifei Hu**:

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**:

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.

Navigation

[0] Message Index

Go to full version