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Verifying That A Function Is A Solution To DE

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello, I am suppose to verify that the indicated function
    [itex]y = \phi (x)[/itex] is an explicit solution of the given first-order
    differential equation. Then I am suppose to consider [itex]\phi[/itex] simply as a function, giving its domain; and then I am suppose to consider it as a solution, giving at least one interval of definition.

    The differential equation: [itex]y' = 25 + y^2[/itex]

    The possible solution: [itex]y = 5 \tan 5x[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I was able to determine the domain to be [itex]\displaystyle ... \cup (\frac{(2k -1) \pi)}{10},\frac{(2k +1) \pi)}{10}) \cup (\frac{(2k +1) \pi)}{10},\frac{(2k +3) \pi)}{10}) \cup (\frac{(2k -3) \pi)}{10},\frac{(2k +5) \pi)}{10}) \cup ... [/itex]

    And I was able to show that the function satisfied the DE, thus being solution:

    [itex]\displaystyle \frac{d}{dx} [5 \tan 5x] = 25 + (5 \tan 5x)^2[/itex]

    [itex]25 \sec^2 5x = 25(1 + \tan^2 5x)[/itex]

    [itex]25 \sec^2 5x =25 \sec^2 5x [/itex], which is a true statement.

    What I am unsure of is, what should the interval of solution be? Does it have to in any way reflect the domain restrictions of the sec function, or only the tan?
     
  2. jcsd
  3. Jan 27, 2014 #2

    LCKurtz

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    Since you need "at least one interval", I would suggest just giving the largest open interval about ##x=0## such that neither the tangent nor secant has a singularity.
     
  4. Jan 27, 2014 #3
    Okay, so I am suppose to find an interval for x such that differential equation and its solution are defined; that is, I have to take into account both tan and sec. I wasn't sure if this was the case or not. Thank you, LCkurtz.
     
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