Question

Pls help:)

Answer 1

Solve in series the Bessel's equation of order 2, \(x^2y"+xy'+(x^2-4)y=0\).

Answer 2

Given \(x^2y"+xy'+(x^2-4)y=0\)-(1)
Let the series solution of (1) be
\(y=\sum_{m=0}^\infty c_mx^{k+m},c_0\ne0\)-(2)
\(.^.. y'=\sum_{m=0}^\infty c_m(k+m)x^{k+m-1},y"=\sum_{m=0}^\infty c_m(k+m)(k+m-1)x^{k+m-2}\)-(3)
Putting the above values of y, y', y" into (1) gives
\(x^2\sum_{m=0}^\infty c_m(k+m)(k+m-1)x^{k+m-2}+x\sum_{m=0}^\infty c_m(k+m)x^{k+m-1}+x^2\sum_{m=0}^\infty c_mx^{k+m}-4\sum_{m=0}^\infty c_mx^{k+m}=0\)
or
\(\sum_{m=0}^\infty c_m{(k+m)(k+m-1)+(k+m)-4}x^{k+m}+\sum_{m=0}^\infty c_mx^{k+m+2}=0\)
or
\(\sum_{m=0}^\infty c_m{(k+m)^2-4)}x^{k+m}+\sum_{m=0}^\infty c_mx^{k+m+2}=0\)
or
\(\sum_{m=0}^\infty c_m(k+m+2)(k+m-2)x^{k+m}+\sum_{m=0}^\infty c_mx^{k+m+2}=0\)-(4)
Equating to zero the coefficient of the smallest power of x, namely \(x^k\), gives the indicial equation
\(c_0(k+2)(k-2)=0~or~(k+2)(k-2)=0~[^..^.c_0\ne0]\)
This gives k=2 and k=-2. These are unequal and differ by an integer. For the recurrence relation, we equate to zero the coefficient of x^{k+m} and get
\(c_m(k+m+2)(k+m+2)+c_{m-2}=0\)
\(.^..~~c_m=\frac{1}{-(k+m+2)(k+m-2)}c_{m-2}\).-(5)
To determine \(c_1\), we equate to zero the coefficient of \(x^{k+1}\) and get \(c_1(k+3)(k-1)=0\) giving \(c_1=0\) for both the roots k=2 and k=-2 of the indicial equation. Now using \(c_1=0\) and (5), we get
\(c_1=c_3=c_5=c_7=...=0\).-(6)
Next, putting m=2,4,6,... in (5) and simplifying, we have
\(c_2=\frac{-c_0}{k(k+4)}\)
\(c_4=\frac{-c_2}{(k+2)(k+6)}\)
\(=\frac{c_0}{k(k+2)(k+4)(k+6)}\),
\(c_6=\frac{-c_4}{(k+4)(k+8)}\)
\(=\frac{-c_0}{k(k+2)(k+4)^2(k+6)(k+8)}\)
and so on. Putting these values in (2), i.e.,
\(y=x^k(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6+c_7x^7+...)\), we get
\(y=c_0x^k{1-\frac{x^2}{k(k+4)}+\frac{x^4}{k(k+2)(k+4)(k+6)}-\frac{x^6}{k(k+2)(k+4)^2(k+6)(k+8)}+...}\)-(7)
Putting k=2 in (7) yields
\(y=c_0x^2{1-\frac{x^2}{2.6}+\frac{x^4}{2.4.8.6}-\frac{x^6}{k(k+4)^2(k+6)(k+8)}+...}\)-(8)
Next, if we put k=-2 in (7), the coefficients of \(x^4\) , \(x^6\), .... become infinite. To get rid of this difficulty, we put \(c_0=d_0(k+2)\) in (7) and obtain modified solution as
\(y=d_0x^k{(k+2)-\frac{(k+2)x^2}{k(k+4)}-\frac{x^4}{k(k+4)(k+6)}-\frac{x^6}{k(k+4)^2(k+6)(k+8)}+...}\)-(9)
Putting k=-2 and replacing \(d_0\) by a in (9) gives
\(y=ax^{-2}(0-0.x^2+\frac{x^4}{(-2)(2)(4)}-\frac{x^6}{(-2)(2)^2(4)(6)}+...)\)
or
\(y=\frac{-ax^2}{16}{1-\frac{x^2}{2.6}+\frac{x^4}{2.4.6.8}-...}=au\),say-(10)
Now,
(8) and (10)=>
\(u=-\frac{x^2}[16}(1-{x^2}{2.6}+{x^4}{2.4.8.6}-...}=-\frac{w}{16}\)
showing that w and u are dependent solutions. Hence we must find one more independent solution in order to obtain the required general solution.
To get another independent solution, substituting (9) into the L.H.S. of (1) and simplifying, we find
\(x^2y"+xy'+(x^2-4)y=d_0(k-2)(k+2)2x^k\)

Answer 3

*\(u=-\frac{x^2}{16}(1-{x^2}{2.6}+\frac{x^4}{2.4.8.6}-...)=-\frac{w}{16}\)

Answer 4

I dnt understand hw the last step comes. can someone pls explain me?

Answer 5

I don't know if this would help you, but in case,

http://www.cs.sjsu.edu/~beeson/courses/Ma133a/ch5.8.pdf

Answer 6

@wio

Answer 7

@John_ES thanx a lot for sharing that pdf file with me I went through it but am sorry i cudnt find what I was looking for.

Answer 8

Ok, try thie other one. May be you would find something usefull.

http://www.unc.edu/~metcalfe/teaching/math524f08/hw/Assignment12.pdf
http://www.math.utah.edu/~gustafso/s2013/3150/asmar-def-theorem-example/ch4AsmarPDF/asmar2004Ch4.7.pdf

Answer 9

And this one,

http://www.solitaryroad.com/c660.html

Answer 10

I am not familiar with this sort of thing, but I have some OCD kicking in.

You use therefore quite a bit, you can just use the code for it.
\(\backslash\text{therefore}=~\therefore\)

Answer 11

ok uhm holy pellet this is a huge math problem I'm starting to have second thoughts about taking a math related job haha xD

Answer 12

wat part u dint got ?