Find the streamlines, particle paths and streaklines when $$u=xe^{2t-z}, \, \, \, v=ye^{2t-z}, \, \, \, w=ze^{2t-z}$$ What is the track of the particle passing through $(1,1,0)$ at time $t = 0$?
To find streamlines we integrate: $$\frac{dx}{u}=\frac{dy}{v}=\frac{dz}{w}$$ but I am confused how to do this because I am not used to having three equalities. I get $$\ln |x|=\ln |y|=z/2 +C$$ which is I don't think is correct.
[The below is a paraphrase of my answer to this question.]
A streamline is a curve which exists in its entirety for every time $t$, but changes with time. To calculate a streamline, we first fix time $t = t_*$. A streamline is a contour of the fixed velocity field. Such a contour line is everywhere tangent to this fixed vector field. To find a parametric representation for such a streamline, denoted by $(x_s(\sigma),y_s(\sigma),z_s(\sigma))$, the fact that the streamline is everywhere tangent to the fixed vector field means that \begin{align} \frac{\text{d} x_s}{\text{d} \sigma} &= x_s \,e^{2t_* - z_s}, \\ \frac{\text{d} y_s}{\text{d} \sigma} &= y_s \,e^{2t_* - z_s}, \\ \frac{\text{d} z_s}{\text{d} \sigma} &= z_s \,e^{2t_* - z_s}. \end{align} We can fix our parametrisation by choosing $(x_s(0),y_s(0),z_s(0)) = (1,1,0)$. You can solve this dynamical system, where the parameter $\sigma$ plays the role of the 'time variable', and obtain the requested stream line.
In particular, I would advise to solve the equation for $z_s(\sigma)$ first, taking into account the 'initial condition' $z_s(0) = 0$. I'm sure you'll have no problems solving the ODEs for $x_s$ and $y_s$ after that.