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Variation of Parameters

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data
    $$y'' - 2y = x + 1$$


    2. Relevant equations
    $$ y_{o} = Ae^{√(2)x} + Be^{-√(2)x} $$
    $$ v_{1}'e^{√(2)x} + v_{2}'e^{-√(2)x}\equiv 0 $$
    $$ √(2)v_{1}'e^{√(2)x}-√(2) - v_{2}'e^{-√(2)x} = x + 1 $$


    3. The attempt at a solution

    $$ v_{2}' = \frac{x+1}{-2√(2)e^{-√(2)x}} $$
    $$ v_{2} = (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x} $$

    $$v_{1}' = (\frac{x+1}{-2√(2)e^{-√(2)x}})(e^{-2√(2)x})$$
    $$v_{1} = (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x}$$

    $$ y = Ae^{√(2)x} + Be^{-√(2)x} + (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x} + (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x} $$

    Where's my error? The correct solution is $$ y = Ae^{√(2)x} + Be^{-√(2)x} - \frac{x}{2} - \frac{1}{2}$$
     
  2. jcsd
  3. Mar 1, 2014 #2
    1. The problem statement, all variables and given/known data
    $$y'' - 2y = x + 1$$


    2. Relevant equations
    $$ y_{h} = Ae^{√(2)x} + Be^{-√(2)x} $$
    $$ v_{1}'e^{√(2)x} + v_{2}'e^{-√(2)x}\equiv 0 $$
    $$ √(2)v_{1}'e^{√(2)x}-√(2) - v_{2}'e^{-√(2)x} = x + 1 $$


    3. The attempt at a solution

    $$ v_{2}' = \frac{x+1}{-2√(2)e^{-√(2)x}} $$
    $$ v_{2} = (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x} $$

    $$v_{1}' = (\frac{x+1}{-2√(2)e^{-√(2)x}})(e^{-2√(2)x})$$
    $$v_{1} = (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x}$$

    $$ y = Ae^{√(2)x} + Be^{-√(2)x} + (\frac{-x}{4}-\frac{√2 + 2}{8})*e^{-√(2)x} + (\frac{-x}{4}+\frac{√2}{8}-\frac{1}{4})*e^{√(2)x} $$

    Where's my error? The correct solution is $$ y = Ae^{√(2)x} + Be^{-√(2)x} - \frac{x}{2} - \frac{1}{2}$$
     
    Last edited: Mar 1, 2014
  4. Mar 1, 2014 #3
    Since I am letting $$ y(x) = v_{1}y_{1} + v_{2}y_{2} $$ my error is that I did not multiply v_{1} and v_{2} by the associated homogeneous solutions, correct?
    $$y_{o}(x) =(\frac{-xe^{√(2)x}}{4}+\frac{√2e^{√(2)x}}{8}-\frac{e^{√(2)x}}{4})*e^{-√(2)x} + (\frac{-xe^{-√(2)x}}{4}-\frac{(√2 + 2)e^{-√(2)x}}{8})*e^{√(2)x} $$
    $$= -(\frac{x}{2}+\frac{1}{2}) $$

    $$ y(x) = Ae^{√(2)x} + Be^{-√(2)x} -(\frac{x}{2}+\frac{1}{2}) $$
     
    Last edited: Mar 1, 2014
  5. Mar 1, 2014 #4

    vela

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    The particular solution is ##y_p = u_1 y_1 + u_2 y_2##. You forgot the y's.
     
  6. Mar 1, 2014 #5

    Dick

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    Variation of parameters is overkill for a problem like this. You should easily be able to find a particular solution to your original ODE by inspection. But your problem is that ##v_1+v_2## isn't part of your solution. You have to multiply them by the homogeneous parts. You want ##v_1 e^{\sqrt{2} x}+v_2 e^{-\sqrt{2} x}##. The exponential parts will cancel in the particular solution.
     
  7. Mar 1, 2014 #6

    Dick

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    Exactly, as I said in your other thread. Please don't double post, I assume this was an accident.
     
  8. Mar 1, 2014 #7
    What do you mean by double posting? Did I post two forums or are you referring to me editing what I've posted on my forum?
     
  9. Mar 1, 2014 #8

    Dick

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    I'm referring to the fact there are two threads called "Variation of Parameters" by vanceEE in the "Calculus and Beyond" forum. You are working on one, I was working on the other.
     
  10. Mar 1, 2014 #9
    Please explain what you mean by solving by inspection.
     
  11. Mar 1, 2014 #10
    Sorry, just noticed that. BTW, what did you mean by solving by inspection?
     
  12. Mar 1, 2014 #11

    Dick

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    Guess a form for the particular solution, say y=ax+b, then y''=0. So y''-2y=x+1 gives -2(ax+b)=x+1, so a=b=(-1/2) and y=-x/2-1/2 works. If the inhomogeneous part is a polynomial, guess a polynomal form and try to solve for it. If it works, it's usually a lot easier than the variation of parameters track.
     
  13. Mar 1, 2014 #12

    Dick

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    See other thread.
     
  14. Mar 1, 2014 #13
    Oh, and yes you're right; I'm aware that I could have used an easier method like undetermined constants to find this solution but the question asked me to use variation of parameters to find a solution for the equation.
     
    Last edited: Mar 1, 2014
  15. Mar 1, 2014 #14

    Dick

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    I thought so, just checking that you also saw how easy it really was if you don't have to use variation of parameters.
     
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