Find all integral curves to the vector field $v(x,y) = x^2\, \partial_x + y\, \partial_y$. See which ones are defined for all $t \in \mathbb{R}$

So, I have to find all curves $\gamma(t)$ that satisfy $v(\gamma(t))=\gamma'(t)$. Right? Let's assume that $\gamma(t)=(\gamma_1(t),\gamma_2(t))$. By plugging it in $v$, we get the equations

$$\frac{d\gamma_1}{dt}=\gamma_1^2 \implies \gamma_1=0 \,\,\,\text{ or }\,\,\,\, \frac{d\gamma_1}{\gamma_1^2}=dt\implies\gamma_1(t)=\frac{1}{C_1-t} \,\text{ or }\, \gamma_1(t)=0$$ $$\frac{d\gamma_2}{dt}=\gamma_2 \implies \gamma_2 = C_2e^t$$

So, it seems that all integral curves for $v$, in the absence of any information about the starting point, must be of the form $\gamma(t)=(\frac{1}{C-t},Ke^t)$ or $\gamma(t)=(0,Ke^t)$.

But what about $\gamma_{\star}=(\frac{1}{t},e^t)$? This also seems to be an integral curve but I can't see how it can be found by solving the differential equations.

Where is my mistake?

Edit: I just figured the answer and realized that I had made a stupid calculational mistake. $\gamma_{\star}=(\frac{1}{t},e^t)$ doesn't satisfy $\gamma_1^2 \partial_x + \gamma_2 \partial_y = \gamma_1' \partial_x + \gamma_2' \partial_y$.


Your $\gamma_1$ isn't correct. $$ \frac{\mathrm{d}\gamma_1}{\gamma_1^2}=\mathrm{d}t\implies\gamma_1(t)=\frac{1}{C-t}. $$

  • $\begingroup$ OK. Thanks. I'll fix it. But how does this answer my question? How do you get $(\frac{1}{t},Ke^t)$ as an integral curve? That's an integral curve. Isn't it? $\endgroup$ – stressed out May 23 '19 at 20:48
  • $\begingroup$ $(t^{-1},e^t)$ is not an integral curve. At any point $(c^{-1},e^c)$, the integral curve through that point is $(1/(c-t),e^{c+t})$. $\endgroup$ – user10354138 May 23 '19 at 20:54
  • $\begingroup$ not sure what you mean. The $C$ is there for the finite-time blowup (forward or backward in time). $\endgroup$ – user10354138 May 23 '19 at 21:02

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