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Given that $$(xy^3+x^2y^7)y'=1$$ satisfies the initial condition $y(\frac{1}{4}) = 1$. Then the value of $y'$ when $y=-1$ is:

  • (A) $\frac{4}{3}$.
  • (B) $-\frac{4}{3}$.
  • (C) $\frac{16}{5}$.
  • (D) $-\frac{16}{5}$.

I doubt if it necessary to solve for $y'$ in order to proceed. Please help.

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If $y(1/4) = 1$, and we want to know the value of $y'$ when $y = -1$, then the Intermediate Value Theorem (assuming $y$ is a continuous function of $x$) tells us that for some $x$, $y = 0$. But this is impossible because the left side of your differential equation would be $0$, while the right side is $1$. Are you sure you copied the problem out correctly? – Amit Kumar Gupta Apr 25 '11 at 5:31
Anyways, I've seen these kinds of questions before, I'm sure there's a slick way to solve them and I'd like to see it. I figure I should know it since I'll have to teach it at some point. Naively, I would sketch the vector field $(x,y) \mapsto (1, y')$, sketch out the trajectory that matches the initial condition, and hope this rough qualitative picture gives me enough of a hint to conjecture the right answer. – Amit Kumar Gupta Apr 25 '11 at 5:38
@Amit Kumar Gupta: I've checked the question once again. There is no error in copying. However, it may be observed that y′ is 16/5 when y=1. – Bhaskar Dey Apr 25 '11 at 7:35
Wolfram Alpha solves the equation:… and with hindsight, knowing the answer, you can solve it using $u=y^4+1/x-4$. It looks like the solution can't be continued past the singularity at $y=0$, so it does seem that there's an error in the problem. – joriki Apr 25 '11 at 10:23
@joriki, how would you use that substitution? Also, do you mean $u = (y^4+1)/(x-4)$ or do you mean what you wrote? – Amit Kumar Gupta Apr 25 '11 at 13:50

Sherlock Holmes: "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."

The question is not a mathematical one, it is more an anti-mathematical one, but the idea is kinda cute. Of course as a Differential Equations question, after a little examination it makes no sense. Let us solve it in almost Sherlock Holmes style. (Sherlock did not know much about singularities. Moriarty would not be impressed.)

Put $y=-1$, and substitute the various suggested values of $y'$ into the equation. We end up with four quadratic equations in $x$. Two of them, corresponding to the suggested answers $4/3$ and $16/5$, have no real solution. The one that comes from the suggested answer $-16/5$ gives $x=1/4$ or $x=-5/4$. The $x=1/4$ is discarded because we were told that there $y=1$. The $x=-5/4$ is presumably to be discarded, for reasons mysterious to me. Maybe there is a glimmer of a genuine differential equations reason, since we cannot get information about the situation when $x<0$ from the initial condition (but of course we cannot get information about $y<0$ either).

So what remains, namely (B), must be the truth. We all know that the argument that led to choosing (B) makes no sense. Someone might want to fool around with the numbers and produce a similar joke question that does not bump into singularity issues. Or not.

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Why would you discard $x=-5/4$? I think Zarrax's response below is correct. – Amit Kumar Gupta Apr 25 '11 at 15:35
@Amit Kumar Gupta: I find that solution rather puzzling. The displayed equation seems to assume that $y$ is identically $1$, since the $y^3$ and $y^7$ terms have mysteriously disappeared. The original differential equation can in fact be solved, and the solution is different from the solution of $y'=-1/(x+x^2)$. But in any case the point of my post (and of several of the comments) is that as a mathematical problem, the whole question makes no sense. – André Nicolas Apr 25 '11 at 16:28
@user6312: I found that confusing, too -- I think $y'=-1/(x+x^2)$ wasn't meant as a differential equation, but as an ordinary equation relating the values at a particular point at which $y=1$. I fully agree that the conclusion should be that the problem is ill-posed. – joriki Apr 25 '11 at 18:16
I agree with both of you that the question is very ill-posed, it does indeed make it sound as though there's only one possible value of $y'$. But re-interpreting it to say, "which of the following values of $y'$ is possible when $y=-1$ (assuming $y$ is a partial function of $x$)?" there is only one answer, and Zarrax's reasoning makes sense. As joriki points out, the resulting equation isn't a differential one. I still don't see why you discarded $x = -5/4$. – Amit Kumar Gupta Apr 25 '11 at 20:12
@Amit Kumar Gupta: I was of course not treating the question seriously, the "for reasons mysterious to me" meant that I did not see why it would be discarded. But since presumably the problem source intended for there to be a unique answer, (s)he probably intended (B). And (s)he was wrong in several ways. – André Nicolas Apr 25 '11 at 20:30

As Amit Gupta pointed out, the solution given doesn't ever go outside the upper right quadrant. So in order for the question to make sense, you'd have to be considering an entirely separate solution on the $x < 0$ domain. Thus you might as well just choose any solution that works just to answer the question. In other words, you look any $x < 0$ such that if you set $y(x) = -1$ as your initial condition then $y'(x)$ is any of four choices given. So you're trying to solve $$ y'(x) = {- {1 \over x + x^2}}$$ Plug in your four possible $y'(x)$ into this equation and try to solve for a negative value of $x$. Since $y'(x) = -{16 \over 5}$ solves for $x = -{5 \over 4}$, this qualifies as a solution to your problem.

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Nice, thanks.${}$ – Amit Kumar Gupta Apr 25 '11 at 15:33
Why not $y'=-\frac{4}{3}$ and $x=-\frac{3}{2}$? The question is phrased as if there's meant to be a unique answer. – joriki Apr 25 '11 at 17:10
Then that qualifies as well. The given solution satisfies $y' > {1 \over 2xy^3}$ as you go to the left from $x = {1 \over 4}$, and will go vertical before you hit the $y$-axis. So the question is ill-posed. – Zarrax Apr 25 '11 at 18:02

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