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Initial Value ODE

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem:

    [tex]
    x^{2}{y}'' + x{y}' + y = 0, x>0, y(1)=1, {y}'=2
    [/tex]


    2. Relevant equations

    y=x^m


    3. The attempt at a solution

    [tex]
    x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
    [/tex]

    [tex]
    x^{2}(m(m-1)x^{m-2})+xmx^{m-1} + x^{m}
    [/tex]

    [tex]
    x^{m}(m(m-1) + m + 1)
    [/tex]

    [tex]
    m = \pm i
    [/tex]

    This is the way i was doing it:
    [tex]
    C_1 e^{it} + C_2 e^{-it}
    [/tex]

    [tex]
    C_1(cos(t) + i sin(t)) + C_2(cos(t) - i sin(t))
    [/tex]


    The solution shows:
    [tex]
    C_1 x^{i} + C_2 x^{-i}
    [/tex]
    [tex]
    C_1(cos(ln(x)) + i sin(ln(x))) + C_2(cos(ln(x)) - i sin(ln(x)))
    [/tex]


    With the initial conditions indicate that the solution is correct. Yet the text book shows the form to be:

    [tex]

    Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}

    [/tex]

    not

    [tex]

    Y(x) = C_1 x^{r_1} + C_2 x^{r_2}
    [/tex]

    And if the second form is correct, I have not found an Euler identity that supports the step.
     
  2. jcsd
  3. Dec 15, 2013 #2

    Mark44

    Staff: Mentor

    Your assumption was that y = xm was a solution. Using that assumption, you solved for m and got ±i.
    I'm confused. You say that
    and then you say
    Which of these does your book show as the solution?

    If the solution you found (i.e., y = c1xi + c2x-i) satisfies the differential equation, then it is the general solution. From the initial conditions you can find the constants c1 and c2.
     
  4. Dec 15, 2013 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You have the correct solution. From your findings of two linearly independent solutions
    [tex]y_1(x)=\exp(\mathrm{i} \ln x), \quad y_2(x)=\exp(-\mathrm{i} \ln x).[/tex]
    The general solution is given by
    [tex]y(x)=C_1 y_1(x) + C_2 y_2(x).[/tex]
    You can also choose any other two linearly independent solutions, e.g.,
    [tex]\tilde{y}_1(x)=\frac{1}{2} [y_1(x)+y_2(x)]=\cos(\ln x), \quad \tilde{y}_1(x) = \frac{1}{2 \mathrm{i}} [y_1(x)+y_2(x)]=\sin(\ln x).[/tex]
    The general solution is again given by all linear combinations of these two functions.
     
  5. Dec 15, 2013 #4
    Okay, where does the ln x come from?
     
  6. Dec 15, 2013 #5

    The textbook shows in the examples the e^{rt} while the solution to this problem shows x^r and there is nothing in the text of the chapter that shows this process. I was able to take cos(ln(x) + i sin(ln(x) back to x^i in wolfram but not sure what i am missing where the ln(x) is coming from and the identity for the x^i.
     
  7. Dec 15, 2013 #6

    Mark44

    Staff: Mentor

    xm = (eln(x))m = em*ln(x). This holds as long as x > 0.
     
  8. Dec 15, 2013 #7
    Sure e^{ln(x)} = x, but where did the ln(x) come from. I have went through the chapters in the book and I have not found a single theorem or example that shows this.

    The examples from the chapter show that when you get your zeros, you use e^{rt). In this case in place of t they used ln(x). I just don't see where the ln(x) came from or what basis this was done with this problem.

    Even over at Pauls notes does not show this:
    http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx
     
    Last edited: Dec 15, 2013
  9. Dec 15, 2013 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Perhaps your text or teacher has talked about how the equation ##ax^2y''+bxy' + cy = 0## can be converted to a similar constant coefficent equation ##a\ddot y + b\dot y + cy = 0## by the change of variable ##t=\ln x##. Then you solve the second equation for y as a function of ##t## with your usual ##e^{rt}## method. Once you have that solution you put ##t=\ln x## in and presto, you have the solution to the original problem. The method you describe sounds like that is what they are doing.
     
  10. Dec 16, 2013 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You did not specify what t is. y is function of x. You assumed the solutions in form [tex]y=x^m [/tex] and you got

    [tex]
    m = \pm i
    [/tex]

    The solutions are y1=xi and y2=x-i, the general solution is y=C1xi+C2x-i.

    [tex]x=e^{\ln(x)}[/tex].

    With that, you can write the solution also in the form
    [tex]
    y=C_1 e^{i\ln(x)}+ C_2 e^{-i\ln(x)}
    [/tex]


    ehild
     
    Last edited: Dec 16, 2013
  11. Dec 16, 2013 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I noticed a typo after it was too late to edit. The equation ##a\ddot y + b\dot y + cy = 0## should have been ##A\ddot y + B\dot y + Cy = 0##. The transformed equation has different constants than the original.
     
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