MHB What are the integral curves of vector field V?

[Aryth1]

My problem is this:

Find the integral curves of $\textbf{V} = (log(y+z),1,-1)$.

I first set up the system:

[math]\frac{dx}{log(y+z)} = \frac{dy}{1} = \frac{dz}{-1}[/math]

I have two find two curves, $u_1$ and $u_2$ that work as integral curves.

The first, and most obvious, function is $u_1(x,y,z) = y + z$. But I'm having trouble finding $u_2$. Any help is greatly appreciated!

 

[Euge]

My problem is this:

Find the integral curves of $\textbf{V} = (log(y+z),1,-1)$.

I first set up the system:

[math]\frac{dx}{log(y+z)} = \frac{dy}{1} = \frac{dz}{-1}[/math]

I have two find two curves, $u_1$ and $u_2$ that work as integral curves.

The first, and most obvious, function is $u_1(x,y,z) = y + z$. But I'm having trouble finding $u_2$. Any help is greatly appreciated!

Hi Aryth,

An integral curve through $\textbf{V}$ is a function of one variable, not three variables. Let's describe it by $\textbf{r}(t) = (x(t), y(t), z(t))$. Solving the system you have above leads to a general solution

$\displaystyle \textbf{r}(t) = (t\log(C_1 +C_2) + B, t + C_1, -t + C_2)$,

where $B, C_1, C_2$ are all constants such that $C_1 + C_2 > 0$. Now you can choose values for the constants to find two integral curves through $\textbf{V}$.