# Finding all integral curves to $x^2 \partial_x + y \partial_y$

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. $$\frac{\mathrm{d}\gamma_1}{\gamma_1^2}=\mathrm{d}t\implies\gamma_1(t)=\frac{1}{C-t}.$$
• 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? – stressed out May 23 '19 at 20:48
• $(t^{-1},e^t)$ is not an integral curve. At any point $(c^{-1},e^c)$, the integral curve through that point is $(1/(c-t),e^{c+t})$. – user10354138 May 23 '19 at 20:54
• not sure what you mean. The $C$ is there for the finite-time blowup (forward or backward in time). – user10354138 May 23 '19 at 21:02