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Homework Help: DE and Complex numbers

  1. Feb 17, 2005 #1
    I believe that I have worked the problem correctly to the point where I am, however I am not sure how to incorporate complex numbers into my answer?

    The problem is :Make the substitution v = ln x to solve 4x^2 * y" + 8xy' - 3y = 0. Where " represents double prime, and ' represents prime.

    This is my work so far:
    v = ln x dv/dx = 1/x

    dy/dx = dy/dv * dv/dx
    dy/dx = dy/dv * 1/x

    d"y/dx" = d/dx [dy/dv * 1/x] - dy/dv * 1/x^2
    = 1/x * d/dx[dy/dv] - dy/dv * 1/x^2
    = 1/x^2 * d"y/dv" - dy/dv * 1/x^2

    Then plugging into the original equation:

    4x^2[t/x^2 * d"y/dv" - dy/dv * 1/x^2] + 8x[dy/dv * 1/x] - 3y = 0
    which can be broken down to
    4d"y/dv" + 4dy/dv - 3y = 0

    Substituting in R, i get 4r^2 + 4r - 3 = 0
    This does not readily look factorable to me so I use the quadratic equation.
    [-4 +- Square Root of (16 - 48)]/8

    Here is where I run into the problem, does that turn into [-2 +- Square root of (8i)] / 4

    It has been a while since I have dealt with complex numbers and I do not recall how to manipulate them.

    If that is the correct equation, can anyone tell me how to finish the problem?

    Thanks,

    Josh
     
  2. jcsd
  3. Feb 17, 2005 #2

    dextercioby

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    Sure,use Euler's identity:
    [tex]e^{i\theta}\equiv \cos\theta+i\sin\theta [/tex]

    for the cmplex exponential part.

    Daniel.
     
  4. Feb 17, 2005 #3
    So would I be correct in saying that :

    y = e^((-1/2)*x)*[(C1Cos((Square root of 2)/2)*X) + (C2Sin((Square root of 2)/2)*X)]

    is correct?

    Where C1 represents some constant and C2 represents some other constant.
     
  5. Feb 17, 2005 #4

    dextercioby

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    Not "x",but "v"...So in terms of "x",u should have some sine & cosine of natural logarithms...

    Daniel.
     
  6. Feb 17, 2005 #5

    ehild

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    You've made a little mistake, it is 16+48 under the square root. You know, (-b+-sqrt(b^2-4ac))/2a. is the solution of the equation ax^2 + bx +c =0.

    ehild
     
  7. Feb 17, 2005 #6
    Wow! Thanks for the help guys!

    And thanks for catching my mistake ehild! I guess that is why I shouldnt work on these problems at the wee hours of the morning.
     
  8. Feb 17, 2005 #7

    dextercioby

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    Ups,i didn't follow your calculations...Just commented on the final result.

    Daniel.

    P.S.So you have a combination of sinh and cosh...
     
  9. Feb 17, 2005 #8

    saltydog

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    This is a Euler equation in the standard form:

    [tex]ax^my^{(m)}+bx^{(m-1)}y^{(m-1)}+...+ky=0[/tex]

    The general solution is of the form: [itex]y(x)=\sum_i^m c_ix^{(m_i)}[/itex]
    where [itex] m_i[/itex] are the roots of the polynomial when [itex]x^m[/itex] is substituted into the ODE.

    The polynomial equation for the ODE of the original posts is:

    [tex]4m^2+4m-3=0[/tex]

    Thus, the solution is seen to be:

    [tex] y(x)=c_1x^{\frac{1}{2}}+c_2x^{\frac{-3}{2}}[/tex]

    Can you attempt to solve the following:

    [tex] 4x^2y''+8xy'+3y=0[/tex]

    This produces complex roots which necessarily involve terms of the form:
    [itex]\cos(\ln[u(x)])[/itex] and [itex]\sin(\ln[v(x)])[/tex] in which [itex]u(x)[/itex] and [itex]v(x)[/itex] are powers of x. Just absorb the imaginary i into the arbritrary constants to get a real-valued function. Although I believe the function is valid only for x>0 because of the logarithm involved. This is what I got:
    [tex]
    y[x]={c_1}\multsp \frac{\cos \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}+\\
    \noalign{\vspace{2.125ex}}
    \hspace{2.em} {c_2}\multsp \frac{\sin \big[\ln \big[{x^{\frac{1}{{\sqrt{2}}}}}\big]\big]}{{\sqrt{x}}}
    [/tex]
     
    Last edited: Feb 17, 2005
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