Vector field and differential equation We consider from $\mathbb{R}^2$ \ $\left \{ 0 \right \}$ the vector field 
$$X(x,y)=\left ( \dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2} \right )$$
How to show that the differential equation $\dot{\gamma }(t)=X(\gamma (t))$ to any initial condition $p \in \mathbb{R}^2$ \ $\left \{ 0 \right \}$ is a curve solution to $\gamma: [0,+\infty) \to \mathbb{R}^2 $ \ $\left \{ 0 \right \}$ with $\gamma(0)=p$. And is this also true if we replace X by -X?
 A: Note that $X$ has the form $X(x,y) = \varphi(x,y) Y(x,y)$ with $\varphi(x,y) = \frac{1}{x^2+y^2}$ and $Y(x,y) = (x,y)$ and the solutions to the system $\dot{\alpha}(t) = Y(\alpha(t))$ are radial line segments. Hence, the solutions of $\dot{\gamma}(t) = X(\gamma(t))$ will also be radial line segments with a different parametrization. Fix $p = (x_0,y_0) \neq (0,0) \in \mathbb{R}^2$ and write $r^2 = x_0^2 + y_0^2$. If we write
$$ \gamma(t) = (x_0, y_0) + \psi(t) \left( \frac{x_0}{r^2}, \frac{y_0}{r^2} \right) = \left(1 + \frac{\psi(t)}{r^2} \right) (x_0, y_0) $$
then
$$ \dot{\gamma}(t) = \frac{\psi'(t)}{r^2} \left( x_0, y_0 \right), \\
X(\gamma(t)) = \frac{1}{1 + \frac{\psi(t)}{r^2} } \left( \frac{x_0}{x_0^2 + y_0^2}, \frac{y_0}{x_0^2 + y_0^2} \right) = 
\frac{1}{r^2 + \psi(t)} \left( x_0, y_0 \right) $$
and so in order for $\gamma$ to be an integral curve of $X$ satisfying $\gamma(0) = (x_0,y_0)$, the function $\psi$ must satisfy
$$ \psi(0) = 0, \frac{\psi'(t)}{r^2} = \frac{1}{r^2 + \psi(t)}. $$
This is an ODE for $\psi$ whose solution is given by
$$ \psi(t) = -r^2 + r\sqrt{r^2 + 2t} = r(\sqrt{r^2 + 2t} - r) $$
leading to
$$ \gamma(t) = \sqrt{1 + \frac{2t}{r^2}} \left( x_0, y_0 \right).$$
