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Differential equation

  1. Mar 16, 2019 #1
    1. The problem statement, all variables and given/known data
    asdf.JPG

    2. Relevant equations
    euler
    ##e^{ix} = cos(x) + i*sin(x)##
    ##e^{-ix} = cos(x) - i*sin(x)##

    3. The attempt at a solution
    I'm starting with differential equations and I'm trying to understand this solution including complex numbers:
    First we determine the zeros. I understand that part. we get ##\lambda_1 = -i## and ##\lambda_2 = i## both with multiplicity 2.
    So we get ##y_1(x)=e^{ix}## , ##y_2(x)=x*e^{ix}## , ##y_3(x)=e^{-ix}## and ##y_4(x)=x*e^{-ix}##. I'm good so far.
    But now we consider the real combinations given by ##y_1(x)=cos(x)## , ##y_2(x)=x*cos(x)##, ##y_3(x)=sin(x)## , ##y_1(x)=x*sin(x)##
    How do you obtain those real combinations? I see that ##Re(e^{ix})=cos(x)## but the sinus parts? And why are we allowed to do that? I can understand that in physics we can focus on the real solution but this is a strict math problem

    Thanks in advance
     
  2. jcsd
  3. Mar 16, 2019 #2

    kuruman

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    Before you consider the real solutions, you need to start with the most general solution and make it match the initial conditions that you are given.
     
  4. Mar 16, 2019 #3
    First, do you understand that any linear combination of solutions to the homogeneous equation is a solution?
    That is, suppose you have ##y_1(x)## such that ##y_1^{(4)} + 2y_1'' + y_1 = 0## and ##y_2(x)## such that ##y_2^{(4)} + 2y_2'' + y_2 = 0##
    Let ##z(x) = ay_1(x) + by_2(x)## where ##a## and ##b## are any numbers, real or complex.

    Then ##z^{(4)} + 2z'' + z = a(y_1^{(4)} + 2y_1'' + y1) + b(y_2^{(4)} + 2y_2'' + y_2) = a*0 + b*0 = 0##.

    From there it's a very simple argument. From the Euler equations you can write the identities
    ##\cos(x) = \frac {e^{ix} + e^{-ix}} {2}##
    ##\sin(x) = \frac {e^{ix} - e^{-ix}} {2i}##

    That is, ##\cos(x)## and ##\sin(x)## are linear combinations of ##e^{ix}## and ##e^{-ix}## (this is why it's important that I emphasized the coefficients in the linear combination can be complex). Thus if the exponentials are solutions, then so are ##\cos## and ##\sin##, though a complex-valued solution might be represented as a combination of ##\cos## and ##\sin## with complex coefficients.

    Alternately, take any linear combination of exponential solutions and just substitute their representation in terms of ##\sin## and ##\cos##. This can always be done.

    Often however, the differential equation represents a physical quantity, so there is the additional restriction that the general solution be real-valued. That restricts the set of allowed complex coefficients for the exponentials, and it's then more convenient to represent the general solution as a linear combination of ##\sin## and ##\cos## with real-valued coefficients.
     
  5. Mar 16, 2019 #4

    epenguin

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    ##y^\left( 4\right) ## is ##\dfrac {d^{4}y}{dx^{4}}## OK? And I will write the operator ##\dfrac {d}{dx}## as ##D##.
    Then your d.e. is

    ##\left( D^{4}+2D^{2}+1\right) y=0##

    Which is

    ##\left( D^{2}+1\right) ^{2}y=0##

    You can surely solve ##\left( D^{2}+1\right) y=0## ? If we call the solution ##y=u(x)## then you just have to solve ##\left( D^{2}+1\right) u=0##
    But you already know the solution to that.

    :devil: Oh. Well up to that point it's simple, but looks like you will get exponentials of exponentials etc. If you do it by stages. Comments invited. But you could just try the trial solution ##y=e^λx## as usual.For that matter you could just try it in the original form of the equation, and the equation for λ will factorise In the same way as above.And looks like double roots will come into it and you will remember there is a certain teaching about that case you can check up about.
     
    Last edited: Mar 17, 2019 at 6:20 AM
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