Sturm-Liouville problem - Partial Differential EquationsI was given the following Sturm-Liouville differential equation problem: x 2 y ″ + 5 x y ′ + λ y = 0 y ( 1 ) = y ( e π ) = 0I displayed the equation in the traditional Sturm-Liouville manner and found that the weight function equals x 3 , but I am having trouble finding the eigenvalues for this problem.We were told in class that because the coefficients of the ODE are functions of x and not constants, we should guess a solution for y in the form of x m and continue from there finding the eigenvalues.Doing that I got a simple quadratic equation involving m: m 2 − 4 m + λ = 0But because it is a quadratic equation where p and q are both non zero I can't seem to find a way to divide the eigenvalues into the common categories which we usually did in class (for λ &lt; 0 , &gt; 0 , = 0), like in this case for instance: m 2 + λ = 0I believe the solution is very simple but just can't find it, help would be much appreciated!

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Sturm-Liouville problem - Partial Differential EquationsI was given the following Sturm-Liouville differential equation problem:      x    2        y    ″    +  5  x      y    ′    +  λ  y  =  0  y  (  1  )  =  y  (      e          π        )  =  0I displayed the equation in the traditional Sturm-Liouville manner and found that the weight function equals       x    3  , but I am having trouble finding the eigenvalues for this problem.We were told in class that because the coefficients of the ODE are functions of x and not constants, we should guess a solution for y in the form of       x    m   and continue from there finding the eigenvalues.Doing that I got a simple quadratic equation involving m:      m    2    −  4  m  +  λ  =  0But because it is a quadratic equation where p and q are both non zero I can&#039;t seem to find a way to divide the eigenvalues into the common categories which we usually did in class (for   λ  &amp;lt;  0  ,  &amp;gt;  0  ,  =  0), like in this case for instance:      m    2    +  λ  =  0I believe the solution is very simple but just can&#039;t find it, help would be much appreciated!

elulamami · Accepted Answer

Step 1If you substitute       x    m   into       x    2        y    ″    +  5  x      y    ′    +  λ  y  =  0, you get  m  (  m  −  1  )  +  5  m  +  λ  =  0        m    2    +  4  m  +  λ  =  0    m  =            −      4      ±              16        −        4        λ              2    =  −  2  ±      4    −    λ  The corresponding solutions are      1          x      2            x                  4        −        λ              ,      1          x      2            x          −              4        −        λ              .To put       x    2        y    ″    +  5  x      y    ′    +  λ  y  =  0 into Sturm-Liouville form, multiply by       x    3  :  −  (      x    5        y    ′        )    ′    =  λ      x    3    y    −      1          x      3        (      x    5        y    ′        )    ′    =  λ  yStep 2This problem is considered in       L                  x        3              2    (  1  ,      e          π        ), where       x    3   is the weight function for the space. The solutions of the above when   λ  =  0 are both in       L                  x        3              2    (  1  ,      e          π        ) because they are regular on (1,eπ). To solve the eigenvalue problem, search for A,B such that  y  (  x  )  =  A      1          x      2            x                  4        −        λ              +  B      1          x      2            x          −              4        −        λ            satisfies  y  (  1  )  =  0  ,      y  (      e          π        )  =  0.The first condition is satisfied by setting   B  =  −  A, which gives                    y        (        x        )                            =                  A                      x            2                                    [                      x                                          4                −                λ                                              −                      x                          −                              4                −                λ                                              ]                                                  =                  A                      x            2                                    [                      e                          i                              λ                −                4                            ln              ⁡              x                                −                      e                          −              i                              λ                −                4                            ln              ⁡              x                                ]                                                  =                  A                      x            2                          2        i        sin        ⁡        (                  λ          −          4                ln        ⁡        x        )            The condition   y  (      e          π        )  =  0 gives the eigenvalue equation  sin  ⁡  (      λ    −    4    π  )  =  0.

I was given the following Sturm-Liouville differential equation problem: x^2y′′+5xy′+lambda y=0, y(1)=y(e^{pi})=0

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