# How to determine phase image

I need to sketch the phase image belonging to the following vector field (I'm sorry, I don't know the exact terms in English, so I have just freely translated them - thanks for sharing the correct terms, if you want to :))

$$F(x,y) = \begin{pmatrix} \frac{1}{2x}\\ yx^2\\ \end{pmatrix}$$

In order to draft the phase image, I have to know the flow $\phi(t, x, y)$ (again, freely translated) of the vector field. Thus, I solved the differential equations $\frac{d}{dt}x(t) = \frac{1}{2x}$ and $\frac{d}{dt} y(t) = yx^2$, which led me to the following result (using initial valus $y(0) = y_0$ and $x(0) = x_0$:

$$\phi(t,x,y) = \begin{pmatrix} \sqrt{t + x^2}\\ ye^{tx^2}\end{pmatrix}$$

My question is: How I am supposed to draft the phase image of this vector flow (it's dependending of three variables, after all) and is it the correct solutiona for the vector flow anyway?

Thanks for your help!

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Welcome to Math.SE! I guess it's better to give an answer late than not to give it at all... The equation for $x$ is autonomous: $\dfrac{dx}{dt}=\dfrac{1}{2x}$, and can be solved by separating variables: $\int 2x\,dx = \int dt$, hence $x^2-x_0^2=t$ and $x=x_0\sqrt{1+t/x_0^2}$. This is almost exactly what you got, except for the sign: $x$ will have the same sign as $x_0$.
Now that $x$ is known, the second equation is $\dfrac{dy}{dt}=y(t+x_0^2)$, which is not autonomous but still separable: $\int \frac{dy}{y}=\int (t+x_0^2)\,dt$. Hence $\log (y/y_0)= t^2/2+x_0^2t$. This leads to the answer $y=y_0\exp(t^2/2+x_0^2t)$ which does not quite agree with yours.
To illustrate the flow of this vector field, we should draw enough trajectories as to make the flow structure evident. "Enough" is subjective. Generally, drawing parametric curves is best done with a computer but here we can do something by hand, because it is easy to eliminate $t$ from the system $$\begin{cases}x=x_0\sqrt{1+t/x_0^2} \\ y=y_0\exp(t^2/2+x_0^2t)\end{cases}$$ via $t=x^2-x_0^2$, hence $y=y_0\exp((x^2-x_0^2)^2/2+x_0^2(x^2-x_0^2))$ which, thankfully, simplifies to $$y=y_0\,\exp\left(\frac{x^4-x_0^4}{2}\right)$$ You get a bunch of curves which move away from the $x$-axis either horizontally (if $y_0=0$) or with rapidly increasing magnitude of $y$.
Thanks a lot! That was very helpful. I made the mistake, that I solved the second equation for $y$ without taking into acount the solution for $x$. Just to make sure (might be a language problem, again): when is a differential equation called 'autonomous'? Do you have an example for an equation that is not autonomous and not separable? –  StringerBell Jan 4 '13 at 12:04
@StringerBell An equation is autonomous if it can be written as $\dfrac{dx}{dt}=g(x)$, without the independent variable $t$ appearing on the right. An equation is separable if it can be written as $\dfrac{dx}{dt}=g(x)h(t)$ for some functions $g$ and $h$. Every autonomous equation is separable because we can think of $g(x)$ as $g(x)h(t)$ where $h(t)\equiv 1$. –  user53153 Jan 4 '13 at 14:44