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Homework Help: Two questions on ODEs

  1. Dec 27, 2005 #1
    Hi all,
    I'd be happy if someone could clarify these two things to me:

    1. While solving linear first-order ODE, I first solve homogenous equation (with the right side equal to 0) and eventually I get to the point (just an example):

    [tex]
    \log |y| = \log C(e^{x} - 1)
    [/tex]

    Now, is it ok to compute [itex]C[/itex] for given non-homogenous equation and then write there two solutions

    [tex]
    y = C(e^{x} - 1)
    [/tex]

    [tex]
    y = -C(e^{x} - 1)\mbox{ ?}
    [/tex]

    Because, you know, both satisfy

    [tex]
    \log |y| = \log C(e^{x} - 1)
    [/tex]

    Anyway, sometimes it gave me correct results (both [itex]C(e^{x} - 1)[/itex] and [itex]-C(e^{x} - 1)[/itex] were solutions) while other times only plus-signed solution was ok.
    2. Solving Bernoulli's ODE, let's say

    [tex]
    y' = 2y + 2x\sqrt{y}
    [/tex]

    and substituting

    [tex]
    z = \sqrt{y}
    [/tex]

    [tex]
    y = z^2
    [/tex]

    [tex]
    y' = 2zz'
    [/tex]

    We get

    [tex]
    2zz' = 2z^2 + 2xz
    [/tex]

    and first thing I do is dividing with [itex]2z[/itex] and so getting the condition [itex]z \neq 0[/itex]
    After some computing, I get the result

    [tex]
    z = Ce^{x} - x - 1
    [/tex]

    and thus

    [tex]
    y = (Ce^{x} - x - 1)^2
    [/tex]

    a) first question
    I'd suppose this expression must not get zeroed because at the beginning we divided the equation with [itex]2z[/itex]. Anyway, even if [itex]C=1[/itex] and [itex]x=0[/itex] (and thus [itex]z = 0 = y[/itex]), the original equation holds true. How it comes?


    b) second question
    Having the substitution above in mind, I know that [itex]z[/itex] itself must be [itex]\geq 0[/itex]. Anyway, if we take [itex]C \in (0, 1)[/itex], then the equation

    [tex]
    Ce^{x} -x - 1 = 0
    [/tex]

    has two roots, [itex]x_0[/itex] and [itex]x_1[/itex]. Then we know that
    for [itex]x \in (x_0, x_1)[/itex]

    [tex]
    Ce^{x} -x - 1 < 0
    [/tex]

    Ok, still remembering that this is [itex]z[/itex] and it must be [itex]\geq 0[/itex], we shouldn't accept this interval for [itex]x[/itex]. However, what's the problem with this solution on this interval? When you put it in the original equation, it's ok I think.

    I can't understand why the interval [itex](x_0,x_1)[/itex] is excluded from the solution (according to our professor).


    Thank you for any suggestions.
     
    Last edited: Dec 27, 2005
  2. jcsd
  3. Dec 27, 2005 #2

    saltydog

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    Homework Helper

    Well, I would've lost money on that one Twoflower. Apparently when you assign:

    [tex]z=y^{1/2}[/tex]

    You implicitly define z[itex]\geq 0[/itex].

    Note that the solution:

    [tex]y(t)=(ce^x-(1+x))^2[/tex]

    does NOT satisfiy the ODE in the interval where z is less than zero. For example, take the plot below for c=0.5 and:

    [tex]y(t)=(ce^x-(1+x))^2[/tex]

    For (approx):

    [tex]x\in (-0.77,1.68)[/tex]

    the curve does not satisfy the ODE. New for me:eek: . Thanks!
     

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  4. Dec 28, 2005 #3

    saltydog

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    I wish to clear up something with this problem from the perspective of existence and uniqueness:

    Consider:

    [tex]y^{'}=f(x,y)=2y+2x\sqrt{y},\quad y(0)=0.25[/tex]

    Now,since f(x,y) and the partial of f with respect to y exists in a bounded region about the point (0,0.25), we can expect a unique solution passing through this point.

    The solution above was determined to be:

    [tex]y(x)=(ce^x-1-x)^2\;\text{with}\;ce^x-1-x\geq 0[/tex]

    Solving for c, I obtain:

    [tex]c=0.5\;\text{and}\;c=1.5[/tex]

    Now, initially this would suggest two solutions satisfy the equation. However, the value of c=0.5 would not satisify the restiction [itex]ce^x-1-x\geq 0[/tex] and so must be discarded. I'm left then with the "unique" solution:

    [tex]y(x)=(1.5e^x-1-x)^2[/tex]
     
  5. Dec 29, 2005 #4
    Thank you Saltydog, now I'm clear about that, it helped me a lot.

    What does this follow from? Is there some theorem saying this?

    Could you clarify those two other questions to me too? Thank you very much.
     
    Last edited: Dec 29, 2005
  6. Dec 29, 2005 #5

    saltydog

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    This is based on Existence and Uniqueness Theorems for ordinary differential equations:

    Consider:

    [tex]y^{'}=f(x,y)[/tex]

    Let T denote the rectangular region defined by:

    [tex]|x-x_0|\leq a\quad |y-y_0|\leq b[/tex]

    with the point [itex](x_0,y_0)[/itex] at its center. Let f and [itex]\frac{\partial f}{\partial y} [/itex] be continuous at each point in T, then there exists an interval[itex]|x-x_0|\leq h[/itex] and a function y(x) such that:

    1. y(x) is a solution to the ODE on the interval [itex]|x-x_0|\leq h[/itex].

    2. y(x) is unique in the sense that if another equation h(x) satisfies the ODE, then y(x)=h(x).

    Note that the interval [itex]|x-x_0|\leq h[/itex] can be very small and nothing is said about uniqueness and existence if f and the partial do not exist at the point x0.

    Thus for a particular IVP, I know if the above is satisfied, I can be guaranteed a unique solution. However, I'm not sure about the case of y=0. Note that for the IVP:

    [tex]y^{'}=2y+2x\sqrt{y};\quad y(0)=0[/tex]

    The partial is indeterminant at the initial condition so uniqueness and existence cannot be guaranteed.

    Not sure I can do better than the above. Perhaps someone in here more knowledgable than I could elaborate on the matter.
     
    Last edited: Dec 29, 2005
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