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Method of reduction of order

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-17_22-21-28.png

    2. Relevant equations


    3. The attempt at a solution
    Should I find the characteristic equation then find the solution y2? Or just integrate the expression of part (a)??

    Part B, just find the derivatives and plug in to the ode
     
  2. jcsd
  3. Oct 18, 2016 #2

    Ssnow

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    yes you can put the expression in the ODE and use the fact that ##y_{1}## is a particular solution ...
     
  4. Oct 18, 2016 #3
    So, I calculate the y2, y2' and y2" in terms of y1, and plug in to ODE (1st line)??
     
  5. Oct 18, 2016 #4

    Ssnow

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    yes, as hint I show you how to derive ##y_{2}(x)## respect ##x##. We observe that we have a product so:

    ##y_{2}'(x)=y_{1}'(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx + y_{1}(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}##

    try for ##y_{2}''## ...
     
  6. Oct 19, 2016 #5
    d/dx ##y_{1}(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}##

    = -1/y1^2 ##e^{-\int p(x)dx}## +1/y p(x) ##e^{-\int p(x)dx}##
     
    Last edited by a moderator: Oct 21, 2016
  7. Oct 19, 2016 #6
    After hving the y, y', y", what should I do?
     
  8. Oct 19, 2016 #7

    Ssnow

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    wait this not ##y_{2}''(x)## this is only a part of the second derivative, you forgot to derive the term ##y_{1}'(x)\int\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx## ...
     
  9. Oct 19, 2016 #8
    Yes, you're right, but is this part correct tho?
     
  10. Oct 19, 2016 #9

    Ssnow

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    yes but there is ##-## instead ##+## in the middle...
     
  11. Oct 19, 2016 #10
    oh right, because of ∫ -p(x) :)
     
  12. Oct 19, 2016 #11
    Then I plug in to the ODE?? but I have 2 unknowns p(x) and q(x) ... what should I do?
     
  13. Oct 20, 2016 #12

    Ssnow

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    If your calculation are correct you must have this:

    ##y_{1}''(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx +y_{1}'(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}+\frac{-p(x)y(x)e^{-\int p(x)dx}-y_{1}'(x)e^{-\int p(x)dx}}{y_{1}^{2}(x)}+##
    ##+p(x)y_{1}'\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx+p(x)\frac{e^{-\int p(x)dx}}{y_{1}(x)}+q(x)y_{1}(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx=0##
     
  14. Oct 20, 2016 #13

    Ssnow

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    that is ##\left[y_{1}''(x)+p(x)y_{1}'(x)+q(x)y_{1}(x)\right]\int\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx=0##, you can simplify and this prove the first part ...
     
  15. Oct 20, 2016 #14

    vela

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    You need to use the fact that ##y_1## is a solution of the ODE.
     
  16. Oct 27, 2016 #15
    thank you
     
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