{VERSION 4 0 "IBM INTEL NT" "4.0" }
{USTYLETAB {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 
0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal
" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }
1 1 0 0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 54 "PHYS 312:\nSOLUTIONS TO \+
PROBLEMS\nSect 4.4\nProblem 4.4:1" }}{PARA 0 "" 0 "" {TEXT -1 565 "Let
 us define $k=\\sqrt\{\\lambda\}$\nThe general solution to\n\\[\\frac
\{d^2\\phi\}\{dx^2\}+k^2\\phi=0\\]\nis\n\\[\\phi=A\\sin(kx)+B\\cos(kx)
\\]\nand\n\\[\\phi^\\prime=Ak\\cos(kx)-Bk\\sin(kx)\\]\n\\[\\phi(0)+\\f
rac\{d\\phi\}\{dx\}|_\{x=0\}=B+Ak\\]\nHence $B=-Ak$ and\n\\[\\phi(a)+
\\frac\{d\\phi\}\{dx\}|_\{x=a\}=(A-Bk)\\sin(ka)+(B+Ak)\\cos(ka)=0\\]\n
The cosine term above vansihes and we get\n\\[A(1+k^2)\\sin(ka)=0\\]\n
or\n\\[k=\\frac\{n\\pi\}\{a\};\\;\\;n=1,3\\cdots\\]\nThe eigenfunction
s can then be written\n\\[\\phi_n=A(\\sin(\\frac\{n\\pi x\}\{a\})-\\fr
ac\{n\\pi\}\{a\}\\cos(\\frac\{n\\pi x\}\{a\}))\\]\nand the eigenvalues
\n\\[\\lambda_n=(\\frac\{n\\pi\}\{a\})^2\\]\n\n" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }