1. Not finding help here? Sign up for a free 30min 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!

Losing a solution of a 1st order ODE?

  1. Mar 10, 2005 #1
    There is an example in a book regarding DEs which I do not understand. Solve the IVP

    [tex] y'=y^2-4, y(0)=-2 [/tex]where t is the independent variable

    We first solve by separation of variables to arrive at the 1-parameter solution.

    [tex] -\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c[/tex]

    Simplifying and expressing the solution explicitly, we find that,

    [tex] y=2\frac{1+ce^{4t}}{1-ce^{4t}} [/tex]

    Taking the initial condition,

    [tex] -2=2\frac{1+c}{1-c} [/tex] which simplifies to,


    They said that the solution is wrong because:

    we can express the DE as, [tex] y'=(y+2)(y-2) [/tex] and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?
    Last edited: Mar 10, 2005
  2. jcsd
  3. Mar 10, 2005 #2


    User Avatar

    Your initial condition y(0)=2 seems to imply that c = 0, not -1=1?
    You can preclude y=-2 simply by noticing that it fails to satisfy the initial condition.
  4. Mar 10, 2005 #3
    sorry, my initial condition is [tex] y(0)=-2[/tex]. Do we ignore [tex]y=-2[/tex] because firstly, it doesn't tell us the value of c, and it gives us the illogical [tex]-1=1[/tex]?

    Also, do we use the other alternative [tex]y=2[/tex] because it is the other "alternative" in [tex]y'=(y+2)(y-2)[/tex]??? If it is so, why?
    Last edited: Mar 10, 2005
  5. Mar 10, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The "separable", non-constant solutions involve a division of both sides of the equation with the expression [tex]y^{2}-4[/tex]
    But, since you can't divide by zero, you have implicitly assumed that [tex]y\neq\pm2[/tex]

    In addition to the non-constant solutions, you've got the constant solutions [tex]y_{1}(t)=2,y_{2}(t)=-2[/tex]
    Last edited: Mar 10, 2005
  6. Mar 10, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    [tex] y(0)=-2 [/tex] is not a valid initial condition.When separating variables YOU ASUMED [tex] y\neq 2;y\neq -2 [/tex] Trying to impose the initial condition to

    [tex] \frac{y-2}{y+2}=Ce^{4t} [/tex]


Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Losing a solution of a 1st order ODE?