aG sin (0) 1 sin“ (0) dg S = -j(j + 1). (19) G sin (0) a0

aG<br>sin (0)<br>1<br>sin“ (0) dg S<br>= -j(j + 1).<br>(19)<br>G<br>sin (0) a0<br>

Extracted text: aG sin (0) 1 sin“ (0) dg S = -j(j + 1). (19) G sin (0) a0

(h) Starting from (19), show that<br>a<br>sin 0-<br>sin ) +2 =<br>G<br>- -j(j+1)G sin? (0)<br>(24)<br>Now, note that the angular differential equation in equation (24) is still<br>a partial differential equation, to convert it into an ordinary differential<br>equation, we will first separate G as follows:<br>G (0, 6) = T (0) Z (4)<br>(25)<br>

Extracted text: (h) Starting from (19), show that a sin 0- sin ) +2 = G - -j(j+1)G sin? (0) (24) Now, note that the angular differential equation in equation (24) is still a partial differential equation, to convert it into an ordinary differential equation, we will first separate G as follows: G (0, 6) = T (0) Z (4) (25)

Jun 04, 2022

SOLUTION.PDF

aG sin (0) 1 sin“ (0) dg S = -j(j + 1). (19) G sin (0) a0

Get Answer To This Question

Related Questions & Answers

Submit New Assignment