 # Problem 1: Find the Wronskian for the following equations: y&q...

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Problem 1: Find the Wronskian for the following equations: y"(x) + k2y = (3) = 0 (4) (1-x2)4"(x)-2xy'(x)+n(n+1)y(x) = (5) Problem 2: We have seen in class that finding the "nice" solution to Bessel's equation is not all that difficult. We'll go through it step by step in the this problem. 1. Assume that the solution of the Bessel equation can be written in the form y(x) = xa 86J. akach k=0 2. Differentiate the above series to obtain expressions for y' (x) and y"(x). 3. Substitute in the ODE (Eq. 4) to show that 00 80 anxath =0 k=0 k=0 80 ao + - + ak + + =0 k=0 (6) 4. In the summation with lower limit from k = 2, put k = l + 2 to change the summation limit to l = 0. The equation you obtain should be: 80 - = 5. Since ao # 0 (if it is zero, the leading term will be xat1 and you'll still have a x term in the rest of the summation - so, you cannot avoid ao), and because n is an integer, the equations you obtain are: a = = (8) = (9) ai al+2 = (10) Clearly, all odd coefficients are zero. We'll pick a = n for the "nice" solution. 6. We can write: 80 + 2 + x21 2 - + - (11) I=1 7. The denominator in the above equation can be modified by re-writing[(n+212 -n²] = (21)(2n+ - = 21) = 21 X 2(n + l). The denominator of the term shown in the above equation becomes: [2(1) X 2(n + 1)] X [2(2) X 2(n + 2)] X 21 X 2 (n + l)] = 2'1! (n+1)! (12) n! 8. Eq. 11 can be simplified to y(x) (13) 9. Since free to choose ao, take it be ao we are to = 80 (14) l!(n + 1)! N 10. Test for convergence for Jn n(x). A series is convergent for a value of x if (-1) (x/2)2(+n 11(n+11) ap- l=0 T1+1 proaches a limit as N 80. This is ensured if < 1 VI M. For > = Ii x2 So, for n > 0, if (l + 1)² 1, i.e., if l > x/2. 4(1+1)(n++++) 11. Generating function for Bessel functions of first kind. Consider the functions exp(xt/2) and exp(-x/(2t)). Their Taylor's series expansions are as follows: 80 = = (15) k (x/2) exp(-x/(2t)) = (16) l! k=0 Let us now multiply the two series and find the coefficient of tn. tn can be obtained by multiplying tn term from the first series with to from second series, or tnt1 from first and t-1 from second, and so on. The coefficient of tn will thus be: 1 80 = = n! 0! (n+1)! 1! + 2)! 2! l=0 The same thing is true for t-n too, except it is (-1)"Jn(x). We can now write: 10.12.1) (17) Put = = - = no. So Eq. 17 can be written as: 80 e ix sin 0 = = + 2 + i2 sin 2nt Equating the real and imaginary parts, we get the following equations: = cos 10.12.2) (19) = 10.12.2) (20) You can find these properties in DLMF - at least some of them can be derived from Eq. 17

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