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. $$ \frac{\mathrm{d}\gamma_1}{\gamma_1^2}=\mathrm{d}t\implies\gamma_1(t)=\frac{1}{C-t}. $$