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Euler's Equations (DiffEq)

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Part 1: Show that Euler's equation has solutions of the form [itex]x^r[/itex]. This can be found by obtaining the characteristic equation [itex] ar(r-1) + br + c = 0 [/itex]
    Part 2: Solve the following Euler equation: [itex] x^{2}y'' + xy' = 0[/itex]

    2. Relevant equations
    Euler's Equation:
    [itex] ax^{2}y'' + bxy' + cy = 0 [/itex]

    3. The attempt at a solution
    For part 1, I didn't have much of a problem, I let [itex] y(x) = x^{r} [/itex]. Therefore:
    [itex] y'(x) = rx^{r-1} [/itex]
    [itex] y''(x) = r(r-1)x^{r-2} [/itex]
    Substituting these values into the Euler equation:
    [itex] ax^{2}(r)(r-1)x^{r-2} + bx(r)x^{r-1} + cx^{r} = 0 [/itex]
    [itex] ax^{2}[r(r-1)x^{r-2}] + bx[rx^{r-1}] + c[x^{r}] = 0 [/itex]
    [itex] a(r^{2}-r)x^{r} + brx^{r} + cx^{r} = 0 [/itex]
    [itex] a(r^{2}-r) + br + c = 0 [/itex]
    [itex] ar(r-1) + br + c [/itex]
    This is the characteristic equation, so that portion of the problem is solved. For part 2, I am not really having a problem formatting the problem, but I am having a problem understanding the given solution. Here is what I did:
    [itex] x^{2}y'' + xy' = 0 [/itex]
    [itex] r(r-1) + r = 0 [/itex]
    [itex] r^{2} - r + r = 0 [/itex]
    [itex] r^{2} = 0 [/itex]
    So the answer has repeated roots with [itex] r = 0 [/itex]. So the solution given in the back of the book is:
    [itex] Y(x) = c_1 + c_{2}ln(x) [/itex]
    I looked around online and verified that this is indeed the answer. However, I am having some trouble understanding where the natural log came from, there is no explanation given in my book and I also can't seem to find a derivation online. Probably I am missing something obvious, but I was hoping someone could help me out and give an explanation, thanks.
     
  2. jcsd
  3. Nov 2, 2014 #2

    RUber

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    I would look at the assumption you used for the derivatives.
    ##y' = x^{r-1} ##
    If r = 0, you either have the trivial solution, or something else that satisfies that assumptions.
     
  4. Nov 2, 2014 #3
    Thanks for the advice...I was wondering, could I also try to derive that equation using reduction of order, similar to how the solution [itex] Y(x) = c_{1}e^{\frac{-bx}{2a}} + c_{2}xe^{\frac{-bx}{2a}} [/itex] is derived?
     
  5. Nov 2, 2014 #4

    LCKurtz

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    Here's an argument that shows you where the ##\ln x## comes from. For simplicity I will call your differential equation ##ax^2y''+bxy'+cy = 0## as ##L(y)=0## where ##L## is the differential operator. When you substitute ##y = x^r## into that equation you got ##L(x^r) = x^rp(r)## where ##p(r)## is the characteristic polynomial. Now consider the case where ##r## is a double root, as you have in your problem. ##r## being a double root means that ##p(r)=0## and ##p'(r)=0##. Now let's differentiate ##L(x^r) = x^rp(r)## with respect to ##r##. If we differentiate the left side, since the order of taking derivatives doesn't matter we get$$
    \frac \partial {\partial r}L(x^r) = L(\frac \partial {\partial r}x^r) = L(x^r\ln x)$$If we differentiate the right side with respect to ##r## we get$$
    (x^r \ln x) p(r) + x^r p'(r) = 0$$since ##r## is a double root. So putting this together we have$$
    L(x^r\ln x) = 0$$which tells you ##x^r\ln x## is a solution of the DE in addition to ##x^r## when ##r## is a double root.
     
  6. Nov 2, 2014 #5
    Fantastic...thank you so much for showing me that. Surprisingly, that argument is hard to find online and I was having some trouble performing it by myself..
     
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