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!

Differential Equation - Initial Value Problem

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the following IVP:

    [tex]\frac{dy}{dt} = \frac{1}{2y+3}[/tex] y(0) = 1

    2. Relevant equations



    3. The attempt at a solution

    [tex]\frac{dy}{dt} = \frac{1}{2y+3}[/tex]
    [tex]\Rightarrow \int 2y + 3 dy= \int dt[/tex]
    [tex]\Rightarrow y^2 + 3y= t +c[/tex] where C is a constant

    Kinda stuck now. Any hints?
     
  2. jcsd
  3. Mar 16, 2009 #2
    If I am understanding the problem correctly, all you have to do is plug in the values for y and t given by the problem text and solve for C.
     
  4. Mar 16, 2009 #3
    dont I need the left side to be y(t)?
     
  5. Mar 16, 2009 #4
    to get your constant c, put in the value of y at t = 0 i.e y(0) = 1,
    we have 1 + 3(0) = 0 + c implies c = 1. Thus you get a quadratic equation, then you can solve for y from there
     
  6. Mar 16, 2009 #5
    wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?
     
  7. Mar 16, 2009 #6
    I do believe you are correct
     
  8. Mar 16, 2009 #7
    so its:
    [tex]y^2 + 3y = t + 4[/tex]
    [tex]y^2 + 3y - 4 = t[/tex]
    [tex](y + 4)(y - 1) = t[/tex]

    stuck again
     
  9. Mar 16, 2009 #8
    Oh, sorry for the mistake. But i know you got what i mean't
     
  10. Mar 16, 2009 #9
    solve [tex]y^2 + 3y -(t + 4) = 0[/tex]
    using the general formula for quadratic equation. you y as a function of t
     
  11. Mar 17, 2009 #10
    So would the general solution be y^2 + 3y = t + c and the unique solution be y^2 + 3y = t + 4?

    I don't think that's it.
     
  12. Mar 17, 2009 #11

    Mark44

    Staff: Mentor

    Does (0, 1) satisfy your unique solution?
    Does your relation satisfy the differential equation? (You'll need to differentiate implicitly if you use y^2 + 3y = t + 4.)

    If the answers to both questions are "yes" you have the unique solution.
     
  13. Mar 18, 2009 #12
    the general solution is y^2 + 3y = t + c because simply it satisfies the differential equation.(check)
    It will be unique if it satisfies the following
    1. dy/dx is continuous in the given interval which contains the initial point

    2. if Lipschitz condition is satisfied.

    y has two roots when you solve the quadratic equation provided the discriminant is not zero. thus the solution of the differential equation is not unique
     
    Last edited: Mar 18, 2009
  14. Mar 18, 2009 #13
    My stupidity knowns no bounds.

    After applying the quadratic formula, I got the following:

    [tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] and [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(t+4)}}{2}[/tex]

    at time t = 0

    [tex]y(t) = \frac{- 3 + \sqrt{9 + 4(0+4)}}{2} = 1[/tex] which satisfies the IVP, while [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(0+4)}}{2} = -4[/tex] which doesn't satisfy the IVP.

    Therefore:

    [tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] is the correct answer.
     
  15. Mar 19, 2009 #14
    Thus, you now know your general solution and the unique solution which satisfies the I.V.P.
     
  16. Mar 19, 2009 #15
    Still can't believe it took me that long to figure it out.

    Thanks for putting up with me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook