# Does x approach some finite value according to this differential equation?

So this question was on a midterm that I wrote recently and I couldn't figure out the answer. We were given the differential equation $$\frac{dy}{dx} = x^2 + y^2$$

The question asked: As $y \rightarrow \infty$, does $x$ approach some finite value $x_0$? Also, we were told to think about it intuitively and without solving the eqution. We have not learned how to solve this sort of equation yet.

I thought about it afterwards and this is the answer I got...

Attempt

The equation of the slope of function y is dependent on both the value of y and the value of x. It is in the form of an equation of a circle.

If $x$ approached some value $x_0$ as $y \rightarrow \infty$, then the slope would gradually approach infinity as $y$ got larger and larger. However, since this is an equation of a circle, the value of the slope is bound by the circle, meaning that it can't approach infinity. Therefore, $x$ does not approach some value $x_0$, and we get that $x \rightarrow \infty$.

Is that correct?

• Every solution $y(\ )$ is increasing hence one can assume that $y(x_1)=y_1$ for some given $x_1$ and some finite $y_1>0$. Then $y'(x)\geqslant y^2(x)$ hence $\left(\frac1{y(x)}\right)'=-\frac{y'(x)}{y^2(x)}\leqslant-1$ hence, for every $x\geqslant x_1$ where the solution is finite, $\frac1{y(x)}-\frac1{y_1}\leqslant x_1-x$, that is, $y(x)\geqslant\frac{y_1}{1+(x_1-x)y_1}$. This proves that $y(x)$ cannot be defined for $x\geqslant x_1+\frac1{y_1}$, that is, that $y(\ )$ explodes at some finite point $x_0\leqslant x_1+\frac1{y_1}$. – Did Oct 28 '16 at 9:29
• @Did: I think you should make that into a solution, it covers a number of broadly useful features in regards to differential inequalities... – copper.hat Oct 28 '16 at 15:56

I don't think this is the case. The value of the slope isn't bound by the circle - it's saying that the value of the slope is ONLY proportional to the distance away from the origin. That is, on every point on the circle $x^2 + y^2 = r^2$, the slope is $r^2$.