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The loss of y = 0 as a solution

  1. Feb 1, 2009 #1

    djeitnstine

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    When solving a Diff. Eq. how do we know that y=0 is another solution lost when we solved it?
     
  2. jcsd
  3. Feb 1, 2009 #2
    You don't know how to check if y = 0 is a solution of a DE?
     
  4. Feb 2, 2009 #3

    Office_Shredder

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    Whatever you divided by is potentially lost as a (when set to zero)solution of your differential equation. So for example, y2 = y*y' you divide by y to get y=y' and you know the solution of that, but since you divided both sides by y, you need to check the case when y=0, since you can't divide by y when y=0
     
  5. Feb 2, 2009 #4

    djeitnstine

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    Thank you for the replies. For NoMoreExams an example here: I did an ODE [tex]\frac{dy}{dx}-y=e^{2x}y^{3}[/tex]

    and solved it using bournulli's method to get

    [tex]y=\sqrt{\frac{2}{-e^{2x} + Ce^{-2x}}}[/tex]

    My professor said y = 0 was lost in that solution.

    So I place y = 0 in the original diff eq. to get

    [tex]\frac{dy}{dx}-0=e^{2x}0^{3}[/tex]

    which is

    [tex]\frac{dy}{dx}=0[/tex]

    which seems wierd
     
  6. Feb 2, 2009 #5
    Remember how you check solutions, LHS = RHS

    Since

    [tex] y = 0 \Rightarrow \frac{dy}{dx} = 0 [/tex]

    So let's look at LHS:

    [tex] \frac{dy}{dx} - y = 0 - 0 = 0 [/tex]

    Now let's look at RHS:

    [tex] e^{2x}y^{3} = e^{2x}0^{3} = 0 [/tex]

    Since [tex] 0 = 0 [/tex], we have shown that [tex] y = 0 [/tex] is a solution to our DE.
     
  7. Feb 2, 2009 #6

    djeitnstine

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    Wow, I can't believe I completely missed that...thanks.
     
  8. Feb 2, 2009 #7
    Never ignore the trivial equilibrium soln.
    They're always the hardest, since you're never really "looking hard" for them.
     
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