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Solving an Euler differential equation

  1. May 26, 2017 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    Solve the differential equation ##(2x+1)^2y'' + (4x+2)y' - 4y = x^2##
    Can someone verify whether my solution is correct?

    2. Relevant equations


    3. The attempt at a solution

    We perform the substitution ##t = \ln|2x+1|##. Then, ##e^t = |2x+1|## and ##x = \pm(e^t -1)/2##

    Without loss of generality, let's solve the equation for ##x = (e^t-1)/2##. (the other case is analogue)

    We find ##\frac{dy}{dx} = 2e^{-t}\frac{dy}{dt}##

    ##\frac{d^2y}{dx^2} = -4e^{-2t}\frac{dy}{dt}+4e^{-2t}\frac{d^2y}{dt^2}##

    Hence, the differential equation becomes:

    ##16y'' - 16y = (e^t-1)^2##

    This is a linear differential equation with constant coefficients:

    The homogeneous is given by:

    ##y_h = c_1e^t + c_2e^{-t}##

    Let's apply variation of the constants to find ##y_p##, the particular solution.

    ##y_p = c_1(t)e^t + c_2(t)e^{-t}##

    Then, we can find the functions ##c_1'## and ##c_2'## as solutions of the system:

    ##\begin{cases}c_1'(t)e^t + c_2'(t)e^{-t} = 0\\c_1'(t)e^t - c_2'(t)e^{-t} = (e^t-1)^2\end{cases}##

    ##\iff c_1'(t) = -c_2'(t) = \frac{(e^t-1)^2}{2e^t}##

    Hence:

    ##c_1(t) = e^t/2 - t - e^{-t}/2 = -c_2(t)## (if we set the constant of integration to ##0##)

    Hence, the solution is given by:

    ##y = y_p + y_h = c_1e^t + c_2e^{-t} -1 + 1/2e^{2t}-te^t + 1/2e^{-2t}+te^{-t}##

    And by substituting back:

    ##y(x) = c_1(2x+1) + c_2\frac{1}{2x+1} + \frac{(2x+1)^2}{2} - \ln(2x+1)(2x+1) - 1 + \frac{1}{2(2x+1)^2} + \frac{\ln(2x+1)}{2x+1}##
     
  2. jcsd
  3. May 26, 2017 #2

    Ray Vickson

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    What is preventing you from putting your formula for ##y(x)## into the DE to test if it works?
     
  4. May 26, 2017 #3

    vela

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    This doesn't look right.
     
  5. May 26, 2017 #4

    Math_QED

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    Thanks for pointing out!
     
  6. May 27, 2017 #5

    epenguin

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    What a pity that last LHS term is –4y instead of 4y, in which case the LHS would have been an exact second derivative. Sure you copied it out right.? :oldsmile:
     
  7. May 27, 2017 #6

    Math_QED

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    I just checked and I made no typo, unfortunately.
     
  8. May 27, 2017 #7

    epenguin

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    Pity they didn't ask good question. I can't see any way to hammer that one into something so nice. :biggrin:
     
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