Can somebody, please, help me? Let $X$ be a vector field on $\mathbb{R}^{2}$, $X=\frac{1}{x}\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$. I was asked to calculate the flow of this vector field..I have to solve a system of differential equations: $\frac{dx}{dt}=\frac{1}{x}$ and $\frac{dy}{dt}=1$; we will have that $y(t)=t+c_{1}$, where $c_{1}$ is a real constant, but for the other one I will have two solutions: $x(t)=\sqrt{2(t+c_{2})}$ or $x(t)=-\sqrt{2(t+c_{2})}$, with $c_{2}$ also a real constant; what is the correct solution? Thank you very much!

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    $\begingroup$ As usual with DEs, these signs and constants are determined by your initial conditions - in this case, the sign of the initial datum $x(0)$. $\endgroup$ – Anthony Carapetis Jun 8 '17 at 8:41
  • $\begingroup$ But my teacher did not tell me what sign $x(0)$ has.. $\endgroup$ – Ramona Ioana Mihu Jun 8 '17 at 8:43
  • $\begingroup$ If the sign is not given, what can I do? $\endgroup$ – Ramona Ioana Mihu Jun 8 '17 at 8:44
  • $\begingroup$ You can give separate formulas for the cases $x(0) > 0$, $x(0) = 0$, $x(0) < 0$...? $\endgroup$ – Andrew D. Hwang Jun 8 '17 at 10:40
  • $\begingroup$ Ok! I did it like that: if $x(0)>0$, we have that the flow is:$F(t,x,y)=(\sqrt{2t+x^{2}}, y+t)$; if $x(0)<0$, we have that the flow is $F(t,x,y)=(-\sqrt{2t+x^{2}},t+y)$. Is this correct? when $x(0)=0$, what is the expression of the flow?with minus or with plus?I dont know that.. Can you, please, help me? thank you! $\endgroup$ – Ramona Ioana Mihu Jun 9 '17 at 7:46

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