A question related to a given differential equation? 
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.
 A: 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.
A: 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. 
