Well we have another homework question from Advanced Calculus that is quite hard to solve, so I thought stackexchange can give me some suggestions on what to do next
Question:
Let $v :\mathbb{R}^3 \to \mathbb{R}^3$ be the vector field given by $v(x,y,z)=[-z,2y,x]$, where $(x, y, z) \in \mathbb{R}^3$.
Find the streamline through the point $(1, 1, 1)$, also known as $p'(t) = v\big(p(t)\big)$, (i.e., $p(0) = (1, 1, 1)$.
My Attempt
First of all, $-dx/z=dy/2y=dz/x$. Hence, we can easily see that $-xdx=zdz$, and if we integrate that, we would get $$-0.5x^2+C_1=0.5z^2+C_2$$ which simply gives $$0.5z^2+0.5x^2=C_1-C_2.$$
We can see that $dz/x=dy/2y$ which gives $2y\ dz=x\,dy$ and if we integrate that again we get $$2y\ z+C_1=y\ x+C_2.$$
I see that we can determine a constant $C_1$ from my point $(1,1,1)$ from that but it doesn't feel right what am I missing and why is it useful that I know these constants.
Instead of finding derivative wrt each other, you need to find it wrt the time.
Solution:
First, we let $p(t) \big( x(t), y(t), z(t)\big)$. From the problem statement, we have \begin{align} -z(t) = x'(t),\\ 2y(t) = y'(t),\\ x (t) = z'(t). \end{align} Using Laplace transformation (or any other fundamental methods), we conclude that \begin{equation} y(t) = C e^{t}. \end{equation} This, together with the fact that $y(0)=1$ concludes that $y(t)=e^{2t}$.
Further, we can derive the following equations using the above ones \begin{equation} x''(t) = -x(t),\\ z''(t)=-z(t). \end{equation} Hence, again using fundamental ODE solutions, we have \begin{equation} x(t)= a_1 \cos (t)+b_1 \sin(t),\\ z(t)= a_2 \cos(t)+b_2 \sin(t). \end{equation} This, together with the facts that $x(0)=z(0)=1$, and $x'(t)=-z(t)$, concludes that \begin{equation} x(t)= \cos(t)-\sin(t),\\ z(t)=\cos(t)+\sin(t), \end{equation} which completes the proof.