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Homework Help: Can I express the maximal solution interval of an ODE as

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm working on trying to first to solve an ODE on the form

    [tex]\sqrt{x(t)}\frac{dx}{dt} = 2\cdot t^2[/tex]

    My task is to find a solution on the form [tex]x(t) =nt^s[/tex]

    3. The attempt at a solution

    I do some seperation and I get that the above is equal to

    [tex]\sqrt{x(t)} dx = 2 \cdot t^2 dt[/tex]

    Which by integrating on both sides of the equality yields

    [tex]\frac{2}{3} x(t)^{3/2} = \frac{2}{3}t^3[/tex]

    I mulitiply by a factor of 3 on both sides and divide by a factor of 2 and arrive at

    and get [tex]x(t)^{3/2} = t^3[/tex]

    which gives the solution

    [tex]ln(x(t)) = 2 \cdot ln(t) \Rightarrow x(t) = t^2[/tex]

    Then the bad not so easy second question. What is the maximum interval on which x(t) is defined?

    According to my textbook I have to equal the two antiderivatives of the original equation to obtain the first endpoint.

    and then take the composite of the invers of the anti-derivative of lhs of the original ODE and the antiderivative of the rhs to find the second endpoint.

    I then obtain an interval [tex]I = [t^2, sign(t)][/tex]

    Have I understood this correctly?
    Last edited: Sep 6, 2010
  2. jcsd
  3. Sep 6, 2010 #2


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    Science Advisor

    I have no idea what you mean by
    Equate what two anti-derivatives? The anti-derivatives of each side of your equation [itex]\sqrt{x}dx= 2t^2 dt[/itex]? You did that to solve the equation.

    Why in the world would you take the composite of the inverses of the two sides? If you have f(y)= g(x), then taking the inverse of f, of both sides, you get [itex]y= f^{-1}(g(x))[/itex] to solve for y but that is the composite of the inverse of the left side with the function on the right side.

    And finally, [itex]I= [t^2, sgn(t)][/itex] means nothing to me. The interval asked for here is the interval between two numbers. Writing two functions together like that is NOT an interval. I would say that because, in the original equation,
    [tex]\sqrt{x}\frac{dx}{dt}= 2t^2[/tex]
    the coefficient of the derivative is only defined and differentiable for x> 0, and that is not true if t= 0, the interval of definition is either [itex](0, \infty)[/itex] or [itex](-\infty, 0)[/itex], depending on whether an "initial condition" is given for t> 0 or t< 0.
  4. Sep 6, 2010 #3
    Hi Mr Hall,

    Thank you for your reply and again I made mistake like a usually do.

    I get that the solution to the above ODE is

    [tex]x(t) = (t^3+\frac{3}{2}c)^{\frac{2}{3}}[/tex]

    If I draw the solution curve on my graphical Calculator I can see this looks to defined on the interval [-3,3].

    Do I then conclude from this Hallsofty that this is the maximum interval for the solution for [tex]k \geq 1[/tex]??
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