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Homework Help: Initial Value problem 1st order ODE

  1. Mar 5, 2014 #1
    1. The problem statement, all variables and given/known data

    I have been trying to follow a solution to a problem I had but do not quite understand the whole thing. I wondered if anybody could clear it up for me.

    Let [tex]a_0[/tex] be the initial value of 'a' for which the transition from one type of behaviour to another occurs.

    The general solution to the ODE is [tex]y=(a+3)e^{\frac{t}{2}}-3e^{\frac{t}{3}}[/tex]

    So to get the critical value between a divergence in behaviour for this equation is when [tex]\frac{dy}{dt}=0[/tex]

    Differentiating the general solution gives:

    [tex]\frac{dy}{dt}= \frac{1}{2}(a+3)e^{\frac{t}{2}}-e^{\frac{t}{3}}[/tex]

    [tex] 0 =\frac{1}{2}(a+3)e^{\frac{t}{2}}-e^{\frac{t}{3}}[/tex]

    Up to here I understand the working. But the next step doesn't make any sense to me where they say "The functions of [tex]e^{\frac{t}{3}}[/tex] and [tex]e^{\frac{t}{2}}[/tex] are not equal to zero..." That's OK but I don't see how it leads to this: "...so each term on the RHS is independently zero"

    According to their solution, this leads to [tex]\frac{1}{2}(a+3)=0[/tex], and so on.

    There are two things I am having difficulty with here:

    • I don't understand what is meant by "each term on the RHS is independently zero". Surely if [tex]e^{\frac{t}{3}}\neq 0[/tex] then [tex](\frac{a+3}{2})e^{\frac{t}{2}}=e^{\frac{t}{3}}[/tex] to make the entire expression zero?

    • This is just a general example and it's not stated that it models anything in reality. But if this was a ODE to represent reality and 't' represented time, wouldn't it make sense to find the critical initial value by differentiating wrt 't' and then find the value of y(0), when t=0?

    I would be very grateful if anyone could give me some pointers!
  2. jcsd
  3. Mar 5, 2014 #2


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    Homework Helper

    I don't understand that either, and I cannot conceive of a meaning of "zero" for which it can be said that [itex]e^{t/3}[/itex] is "zero", although the author does get the correct answer.

    Setting the derivative to zero tells you the value(s) of [itex]t[/itex] at which [itex]y(t)[/itex] has a critical point. You can then try to work out whether, depending on the value of [itex]a[/itex], there are any critical points in [itex]t > 0[/itex] and, if so, whether they are local maxima, minima, or points of inflection by considering the second derivative. That information will tell you whether [itex]y[/itex] is eventually increasing or decreasing.

    Alternatively you can observe that [itex]y(t) = e^{t/2}(3 + a - 3e^{-t/6})[/itex], so that [itex]y \to + \infty[/itex] if [itex]3 + a > 0[/itex] and [itex]y \to -\infty[/itex] if [itex]3 + a \leq 0[/itex], so the critical value is [itex]a = -3[/itex].

    Having found the critical values of the constant of integration by some means, you can then indeed find corresponding critical initial values by substituting them into the general solution and setting [itex]t = 0[/itex]. That's what the author has done here, by the cunning method of writing [itex]Ae^{t/2} - 3e^{t/3}[/itex] as [itex](3 + a)e^{t/2} - 3e^{t/3}[/itex] so that [itex]y(0) = a = A - 3[/itex].

    Whether the ODE is asserted to a be model of any physical system is entirely irrelevant here. Critical values are critical because the mathematics says they are.
  4. Mar 5, 2014 #3
    Thanks a lot pasmith. I'm beginning to understand it now.

    Just in terms of finding the critical value of the constant of integration I think I can see that
    But I'm not too sure about any alternative way to find [tex]a_0[/tex].

    It seems that in all the examples I have done up until now the equilibrium solution has occurred where c=0. I don't know if this is just a co-incidence: if it isn't, it would make finding the critical value for the constant of integration easy: just set it so that the constant of integration =0.
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