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Homework Help: Differential equations (initial-value problem)

  1. Mar 23, 2005 #1
    Consider the initial-value problem

    [tex]y^{\prime} + \frac{2}{3} y = 1 - \frac{1}{2}t,\quad y(0) = y_0[/tex]

    Find the value of [tex]y_0[/tex] for which the solution touches, but does not cross, the [tex]t[/tex]-axis.

    The only problem that I seem to have is finding this particular solution that "touches, but does not cross, the [tex]t[/tex]-axis".

    Any help is highly appreciated.

    First, I use the method of integrating factors.

    [tex]y^{\prime} + \frac{2}{3} y = 1 - \frac{1}{2}t \qquad \mbox{(Standard Form)}[/tex]

    It follows that



    [tex]g(t)=1 - \frac{1}{2}t[/tex]. Thus, we find

    [tex]\mu (t) = \exp \int p(t) \: dt = \exp \frac{2}{3} \int \: dt = e^{2t/3}[/tex]


    [tex]y(t)=\frac{1}{\mu (t)}\left[ \int \mu (t) g (t) \: dt + \mathrm{C} \right][/tex]
    [tex]y(t)=e^{-2t/3}\left[ \int e^{2t/3} \left( 1 - \frac{1}{2}t \right) \: dt + \mathrm{C} \right][/tex]
    [tex]y(t)=e^{-2t/3}\left[ \left( \frac{21}{8} - \frac{3}{4}t \right) e^{2t/3} + \mathrm{C} \right][/tex]
    [tex]y(t)=\frac{21}{8} - \frac{3}{4}t + \mathrm{C} e^{-2t/3}[/tex]

    The initial condition gives

    [tex]y_0 = \frac{21}{8} - \frac{3}{4}(0) + \mathrm{C} e^{-2(0)/3} \Longrightarrow \mathrm{C} = y_0 - \frac{21}{8}[/tex]

    So, we get

    [tex]y(t)=\frac{21}{8} - \frac{3}{4}t + \left( y_0 - \frac{21}{8} \right) e^{-2t/3}[/tex]

    The problem now is finding the appropriate value for [tex]y_0[/tex]. The answer my textbook gives is [tex]y_0\approx -1.642876[/tex], but so far I haven't been able to figure out the way to work backwards from there. Anyhow, I tried to find it by plotting some of the solutions. It turns out that all of them are horizontal lines. So, maybe I just need to find [tex]y_0[/tex] that gives the line [tex]y=0[/tex]. But again, I'm not so sure. Am I on the right track?

  2. jcsd
  3. Mar 23, 2005 #2


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    Homework Helper

    Hello Thiago. The equation is not a horizontal line. Not sure I understand what you mean by that. How about this: if it's continuous and touches the x-axis but does not cross it (and is not y=0), then aren't y and y' zero there? So that's two equations in two unknowns, albeit non-linear ones. It's a start anyway.
  4. Mar 23, 2005 #3
    Yeah... you're right! What was I thinking?!? If it were a line, it'd fit the form y = ax + b. I must have overlooked my y. :rofl:

    Anyway, I see what you mean. I'll use y = 0 and y' = 0 this time.

    Last edited: Mar 23, 2005
  5. Mar 23, 2005 #4
    Hey, thank you so much. That works fine!

    First, we look into [tex]y=0[/tex].

    [tex]\frac{21}{8} - \frac{3}{4}t + \left( y_0 - \frac{21}{8} \right) e^{-2t/3} = 0[/tex]

    Solving it for [tex]y_0[/tex] gives [tex]y_0 = \frac{3}{8} \left( 7 - 7 e^{2t/3} + 2t e^{2t/3} \right)[/tex]. Next, we look into [tex]y^{\prime}=0[/tex].

    [tex]-\frac{3}{4} -\frac{2}{3}\left( y_0 - \frac{21}{8} \right) e^{-2t/3} = 0[/tex]

    Using the [tex]y_0[/tex] found in the last step, and solving for [tex]t[/tex] gives [tex]t=2[/tex]. Finally, we substitute that value back in [tex]y_0[/tex] and get [tex]y_0 = \frac{3}{8}\left( 7 - 3e^{4/3} \right) \approx -1.642876[/tex]

    That's it!
  6. Dec 9, 2010 #5
    this helped me soooo much!! thanks you
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