1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Initial value problem

  1. May 14, 2014 #1


    User Avatar

    1. The problem statement, all variables and given/known data

    The problem is from Walter Gautschi - Numerical Analysis, exercise 5.1.

    Consider the initial value problem

    [itex]\frac{dy}{dx}=\kappa(y+y^3), 0\leq x\leq1; y(0)=s[/itex]

    where [itex]\kappa > 0[/itex] (in fact, [itex]\kappa >> 1[/itex]) and s > 0. Under what conditions on s does the solution [itex]y(x) = y(x;s)[/itex] exist on the whole interval [0,1]? {Hint: find y explicitly.}

    3. The attempt at a solution

    Following the hint, I tried solving it as a separable differential equation:

    [itex]\frac{dy}{y+y^3} = \kappa dx[/itex]

    (Using wolframalpha here)

    [itex]log(y) - \frac{1}{2}log(y^2+1) = \kappa x[/itex]

    [itex]10^{log(y)-\frac{1}{2}log(y^2+1)} = 10^{\kappa x}[/itex]

    Ending up with

    [itex]\frac{y}{2} + \frac{1}{2y} = 10^{-\kappa x}[/itex]

    Not sure how to solve this, so used wolframalpha again and got:

    [itex]y = 10^{-\kappa x} \pm \sqrt{10^{-2 \kappa x} - 1} [/itex]

    Evidently [itex]y(0) = 1[/itex], which means that s has to be equal to 1.

    This doesn't seem right to me. Especially considering the work needed to find y explicitly, seeing as how this is not a course in differential equations. So what I am missing here?
  2. jcsd
  3. May 14, 2014 #2


    User Avatar
    Homework Helper

    I see no constant of integration; the right hand side should be [itex]\kappa x + C[/itex].
  4. May 16, 2014 #3


    User Avatar

    Thank you, and right you are; that was quite sloppy of me. So starting then at

    [itex]\frac{y}{2}+\frac{1}{2y}=C10^{\kappa x}[/itex]


    [itex]y = C10^{-\kappa x} \pm \sqrt{ C^{2}10^{-2 \kappa x} - 1 }[/itex]


    [itex] y(0) = s = C \pm \sqrt{C^2-1}[/itex]

    Which imposes a constraint on C: [itex] |C| \geq 1 [/itex]

    But they are asking for what conditions on s the solution [itex]y(x) = y(x;s)[/itex] exists on the interval [0,1]? I am having some trouble seeing the connection here, especially in the context of the chapter "Initial Value Problems for ODEs: One-Step Methods". Any thoughts?
  5. May 16, 2014 #4


    User Avatar
    Homework Helper

    [tex]\log(y) - \frac12 \log(y^2 + 1) = \log\left( \frac{y}{\sqrt{1 + y^2}}\right).[/tex] This is not in general equal to [itex]\log(y/2 + 1/(2y))[/itex], which is what your expression implies (and I must apologise for not pointing this out in my first reply).

    Also: in calculus, "log" means "natural logarithm", so if [itex]a = \log b[/itex] then [itex]b = e^a[/itex].
  6. May 16, 2014 #5


    User Avatar

    Ah, that explains the seemingly frivolous interchangeable use of log and ln I've sometimes seen. And yes, I see that I've made en error when dealing with the [itex] \frac{1}{2} [/itex] factor. Well, being more explicit this time and starting from (I will for my own sake switch to using ln now):

    [itex] ln(y) - \frac{1}{2}ln(y^2+1) = \kappa x + C_1 [/itex]

    [itex] ln(y) - ln( \sqrt{y^2+1}) = \kappa x + C_1 [/itex]

    [itex] e^{ln(y) - ln(\sqrt{y^2+1})} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{e^{ln(y)}}{e^{ln(\sqrt{y^2+1})}} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{y}{\sqrt{(y^2+1)}} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{y^2}{y^2+1} = C_3 e^{2\kappa x} [/itex]

    [itex] \frac{y^2+1}{y^2} = C_4 e^{-2\kappa x} [/itex]

    [itex] 1 + y^{-2} = C_4 e^{-2\kappa x} [/itex]

    [itex] y^{-2} = C_4 e^{-2\kappa x} - 1 [/itex]

    [itex] y^2 = \frac{1}{C_4 e^{-2\kappa x} -1} [/itex]

    [itex] y = \pm \frac{1}{\sqrt{C_4 e^{-2\kappa x} -1}} [/itex]


    [itex] y(0) = s = \pm \frac{1}{\sqrt{C_4 - 1}} [/itex]

    and thus [itex] C_4 > 1 [/itex].

    Hopefully the math checks out this time. But I am still not sure how to use this in the context of the exercise. Any ideas?
  7. May 16, 2014 #6


    User Avatar
    Homework Helper

    It was given that s>0. Find the integration constant C4 in terms of s. Than write y(x) using s, and see what should be true for s, so y(x,s) is defined for the whole interval [0,1].

  8. May 16, 2014 #7


    User Avatar

    Perfect, thanks!
  9. May 16, 2014 #8


    User Avatar
    Homework Helper

    You are welcome:smile:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted