2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $x_0\leqslant x_1+\frac1{y_1}$. $\endgroup$ – Did Oct 28 '16 at 9:29
  • $\begingroup$ @Did: I think you should make that into a solution, it covers a number of broadly useful features in regards to differential inequalities... $\endgroup$ – copper.hat Oct 28 '16 at 15:56
1
$\begingroup$

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 $r^2$.

From this picture, you see that as the distance from the origin increases, the slope increases more and more rapidly. So this makes it seem that there ought to be an asymptote.

$\endgroup$
  • $\begingroup$ I see. Oh well. It wasn't worth a lot of marks on the midterm anyway. Thanks $\endgroup$ – The_Questioner Oct 28 '16 at 2:50
  • $\begingroup$ @The_Questioner Perhaps the marks were justified after all, see comment on main. $\endgroup$ – Did Oct 28 '16 at 9:31
  • $\begingroup$ Doesn't your comment agree with mine? Why the downvote (if that was you)? $\endgroup$ – Nitin Oct 28 '16 at 10:14
  • $\begingroup$ @Nitin. I didn't downvote you. $\endgroup$ – The_Questioner Oct 29 '16 at 1:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.