Finding Streamlines: How to solve this ODE? Is there a easier way?

Question

I was given the velocity vector field $\mathbf{\vec{v}}(x, y, z)$ and was asked to find the streamlines $\mathbf{\vec{r}}(t)$ in my exam:

$$\mathbf{\vec{v}}(x, y, z) = -\dfrac yx \hat i + 4x \hat j + 0 \hat k$$

This is what I did:

$$z(t) = \int 0 dt = C_1$$

$$\dfrac{dx}{dt} = -\dfrac {y(t)}{x(t)};\; \dfrac{dy}{dt} = 4x(t)$$

But these two equations are coupled. So, I tried to decouple them like this:

$$y = -x\dfrac{dx}{dt}$$

$$\implies \dfrac{dy}{dt} = -\left(\frac{dx}{dt}\right)^2 - x\dfrac{d^2x}{dt^2}$$

$$\implies x\dfrac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2 + 4x = 0$$

Now, what do I do? Is there a easier way?

Answer 1

In $2$ dimensions you can transform a system $$\dot x=-{y\over x},\quad \dot y=4x$$ directly into a $(x,y)$-ODE for the stream lines: Write $$y'={\dot y\over\dot x}=-{4x^2\over y}$$ and obtain $$(y^2)'=-8x^2\ .$$ It follows that the stream lines are given by $$y^2+{8\over3}x^3=C\ .$$ Of course the information about the time dependence is lost in this process.

Answer 2

setting $$u(x)=\frac{dx(t)}{dt}$$ so we get $$4x+u(x)^2+x\frac{du(x)}{dx}u(x)=0$$ or $$x\frac{du(x)}{dx}u(x)+u(x)^2=-4x$$ and we substitute $$u(x)=v(x)^2$$ with $$\frac{du(x)}{dx}=2v(x)\frac{dv(x)}{dx}$$ and we get $$\frac{dv(x)}{dx}+2\frac{v(x)}{x}=-8$$ with the integrating factorn $$\mu(x)=x^2$$ $$x^2\frac{dv(x)}{dx}+2xv(x)=-8x^2$$ and we have $$\int \frac{d}{dx}(x^2v(x))dx=\int -8x^2dx$$ and we get $$x^2v(x)=-\frac{8}{3}x^3+C_1$$ i think now it is right!