1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Initial Value Problem Question - Differential Equations

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the initial value problem where y(0) = 2.

    2. Relevant equations
    dy/dt = 3 - 5(y^(1/2))

    3. The attempt at a solution
    I tried the separable equation method but when it came time to take the integral of
    1/[3 - 5(y^(1/2)], every solution I got became too complex to solve for y.
    I'm thinking maybe you can just somehow put this into the linear form and use Leibniz's method by multiplying the equation by u(t) but I don't see how I can do this.
    All I need is some info on how to make the integral simpler (so I can eventually solve for y) or how I can change this into linear form.
  2. jcsd
  3. Oct 11, 2008 #2
    [tex]\frac{dy}{dt}=3-5y^{\frac 1 2}[/tex]

    Correct? If so, have you learned these 2 methods: Integrating factor or Variation of parameters?
  4. Oct 11, 2008 #3
    That is correct (the y^1/2 is a square root but I can't really type that).
    I've learned the integrating factor method, I believe, but not variation of parameters.

    Try explaining what you're thinking and I'll see if I understand or recognize what I have learned :)
  5. Oct 11, 2008 #4
    Well you can't easily integrate this problem, so you will need to use the integrator factor.
  6. Oct 11, 2008 #5
    Will you demonstrate how to do this? or get me started? Maybe it's obvious but I can't seem to figure it out.
  7. Oct 11, 2008 #6
    I know how to do the integrating factor method, but not with this problem. How would i get rid of the square root so as to have this in its linear form? (dy/dt + p(t)y = g(t))
  8. Oct 11, 2008 #7


    User Avatar
    Homework Helper

    Separation of variables works here. The expression is also easy to integrate with an appropriate substitution.
  9. Oct 11, 2008 #8
    Any hints as to what that substitution might be? I've tried letting u = 3-5(y^(1/2)) and
    u = y^(1/2) and I've even tried multiplying top and bottom by 3 + 5(y^(1/2)) in order to remove the square root from the bottom. Nothing has worked for me so far...I must be missing something really obvious.
  10. Oct 11, 2008 #9


    User Avatar
    Homework Helper
    Gold Member

    Show us your work for the u=y^1/2 substitution; you should be able to integrate it without much trouble using this sub.
  11. Oct 11, 2008 #10


    User Avatar
    Homework Helper

    Why doesn't this work?
  12. Oct 11, 2008 #11
    Ok. Maybe I just made a simple mathematical error the first time around. Let me give it another try and I'll get back to you.
  13. Oct 11, 2008 #12
    DOH! haha. Ok I see it now. For some reason it didn't work last night (I must have been tired). I was able to fully integrate.
    Thanks so much guys for your help! (and your patience :) )
  14. Oct 11, 2008 #13
    Oh crap it is ... lol, just divide the right and get dt on the right ... sorry!!!
  15. Oct 11, 2008 #14
    Um...ACTUALLY I have one more question. First off, I accidentally made a typo...it's actually 5 - 3(y^(1/2)) NOT 3 - 5(y^(1/2)). Second, after I took the integral of 1/(5-3(y^(1/2))) I got (-10/9)ln l 5 - 3(y^(1/2)) l + (2/9)(5-3(y^(1/2))) = t + C. How would I solve for y in this case? Any hints would be appreciated :)
  16. Oct 11, 2008 #15


    User Avatar
    Homework Helper

    Do you really have to solve in terms of y? Remember that the only thing you're required to do is to solve an initial value problem. You're given the initial condition. So just plug that in to find C. It's not always possible to express a function f(y,t) = g(t) in terms of y only on the left hand side. This has something to do with the implicit function theorem:
  17. Oct 11, 2008 #16
    I don't know if I HAVE to solve in terms of y, but all the examples/problems that we've done in class have been in that form so I really don't know.
    So I take it there is no real way of finding it it in terms of y on the left hand side?
  18. Oct 11, 2008 #17
    Actually, I take that back. There is a second part to the problem. I have to use Euler's method to approximate at several values of t. I take it I need to first solve for y in this IVP problem so I can use it in Euler's method. Any ideas on how I'm supposed to do this?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Initial Value Problem Date
Initial value problem question Feb 17, 2018
Solution to complex valued ODE Feb 15, 2017
Differential Equation Initial Value Problem Feb 14, 2017
Initial Value Problem Jan 22, 2017
Initial Value Problem for (DE) Jan 17, 2017