Sketch the direction field of the differential equation $y'=x+y^2$. Then use it to sketch a solution curve that passes through the point $(0,0)$.
My calculator has given me the direction field and it's the same as Wolframalpha http://www.wolframalpha.com/input/?i=streamplot+%5B%7B1,x%2By%5E2%7D,+%7Bx,-4,2%7D,+%7By,-2,4%7D%5D
I don't know if this graph is right though
My teacher has also asked for a table of values so we can plot the direction field. I don't know what values to put into the table.
Best Answer
$$y'=x+y^2 \quad\text{ with condition }\quad y(0)=0$$ This is an ODE of the Riccati kind. The general solution involves some Bessel functions or the Airy functions. Of course, this isn't what is asked for.
In order to sketch the behavior of the particular solution $y(x)$ with respect to the condition $y(0)=0$ we consider first the behavior close to $0$. Then, $y^2$ is small so that $y'\simeq x$ which leads to $$y\simeq \frac{x^2}{2} \quad\text{small } x$$
For $x$ increasing, $y'=x+y^2>0$ implies $y$ increasing the more and more quickly as $y^2$ becomes the more and more large.
For large values of $y$ :$\quad y'\simeq y^2$ which leads to $$y\simeq \frac{1}{c-x} \quad\text{large } y \text{ and } 0<x<c$$
Thus, $y$ tends to $\infty$ for $x$ tending to a value $c$ which cannot be computed with a so simplified method. Anyways, there is a vertical asymptote.