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Separation of Variables

  1. Mar 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Using separation of variables determine if the solution escapes to infinity in finite time or infinite time?

    [tex]y'(t)=1+\frac{y(t)}{2}[/tex]
    [tex]y(0)=.5[/tex]

    2. Relevant equations
    Knowing how to do separation of variables.


    3. The attempt at a solution
    Here is my attempt, but I get stuck...
    [tex]y'(t)=1+\frac{y(t)}{2}[/tex]
    [tex]y'(t)-\frac{y(t)}{2}=1[/tex]
    [tex]\int_0^t{y'(x)-\frac{y(x)}{2}dx}=\int_0^t{1dx}[/tex]
    The next step I'm not sure of...
    [tex](y(t)-y(0))-(\frac{y(t)^2}{4}-\frac{y(0)^2}{4})=t[/tex]
    [tex]y(t)-\frac{y(t)^2}{4}=t+y(0)-\frac{y(0)^2}{4}[/tex]
    Now solving for [tex]y(t)[/tex] becomes a problem if the above step is correct... I'm sure I'm doing something wrong.
     
    Last edited: Mar 20, 2008
  2. jcsd
  3. Mar 20, 2008 #2
    Sorry about that... I'm learning DiffEq through Mathematica and needless to say, it's poop.

    Anyway, I figured out how to do it.

    [tex]y'(t)=1+\frac{y(t)}{2}[/tex]
    [tex]\frac{dy}{dt}=\frac{2+y(t)}{2}[/tex]
    [tex]dy=\frac{(2+y(t))dt}{2}[/tex]
    [tex]\frac{dy}{y(t)+2}=\frac{dt}{2}[/tex]
    [tex]\int{\frac{dy}{y(t)+2}}=\int{\frac{dt}{2}[/tex]
    [tex]\ln{(y(t)+2)}=\frac{t}{2}+C[/tex]
    [tex]y(t)+2=Ce^{t/2}[/tex]
    [tex]y(t)=Ce^{t/2}-2[/tex]
    [tex]y(0)=.5=Ce^{0/2}-2[/tex]
    [tex]C=2.5[/tex]
    [tex]y(t)=2.5e^{t/2}-2[/tex]
     
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