Ma2aPracLecture24 - Lecture 24 To begin it may be helpful...

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Lecture 24 To begin, it may be helpful to make a few comments on the case of regular singular points before starting the next big topic of the course, namely the Laplace Transform. If x = x 0 is a regular singular point of a linear, second order ODE ( ) y ′′ + p ( x ) y + q ( x ) = 0 , then we look for series solutions just like in the case of an ordinary point, but with a difference, namely we try y = x r n =0 a n ( x x 0 ) n , where r is a constant to be determined, which r need not be an integer, or even a rational number. (When r is not rational, it will be a quadratic irrational.) Assuming that the radius of convergence R (around x 0 ) is pos- itive, we differentiate the series expression term by term (for x satisfying | x x 0 | < R ) to get series expressions for y and y ′′ . Plugging this informa- tion back in the differential equation, we get an identity of the form x r n =0 c n ( x x 0 ) n = 0 , where the coefficients c n of the power series on the left are determined by the coefficients a n of y (and hence of y , y ′′ ) as well as by r, p ( x ) .q ( x ). It follows that for such an identity to hold for all x close to x 0 , we need to have c n = 0 , n 0 . A key difference with the ordinary case occurs when we look at the condition c 0 = 0, because this leads to an equation for r , called the indicial equation associated to ( ), which will be a quadratic equation in r . In this course we will only look at the case when this has two distinct real roots, and moreover assume that the two roots do not differ by a integer. (As usual, it is more subtle when there is a repeated root, and even when the roots are not repeated, if they differ by a positive integer, we will not be able to find 1
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easily a second series solution.) For an r satisfying the indicial equation, we may plug its value in and solve the equations c n = 0 for all n > 0. This results, as in the ordinary case, a formula for the coefficients a n , which can be used to find an explicit series solutions of ( ). Let us illustrate this by solving the following ODE:
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Ma2aPracLecture24 - Lecture 24 To begin it may be helpful...

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