2
$\begingroup$

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.

  • 3
    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?2011-04-25
  • 0
    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.2011-04-25
  • 0
    @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=1.2011-04-25
  • 3
    Wolfram Alpha solves the equation: http://www.wolframalpha.com/input/?i=solve+%28xy%5E3%2Bx%5E2y%5E7%29y%27%3D1 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.2011-04-25
  • 0
    @joriki, how would you use that substitution? Also, do you mean $u = (y^4+1)/(x-4)$ or do you mean what you wrote?2011-04-25
  • 0
    @Amit: I do mean what I wrote :-). Making that substitution leads to $u'(1+4/u)=-1/x^2$, which you can integrate to $u+4\ln u=1/x+c$, which is solved by $u=4W(c'\exp(1/4x))$.2011-04-25

2 Answers 2