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sin 8- + jG+1) sin² (@) = k² (27) sin e. and, (28) Z dộ? Show that equation (27) combined with equation (28) in a certain way gives us (26) (this means that equation (27) along with (28) is completely equivalent to equation(26)). [3] The solution for equation (27) requires power series analysis, which we will not go into; the solution is as follows: T (0) = T;x (0) = n;aP} (cos (0)), (29) here njk are constants that can be specified with integers j and k; P is a special function called Legendre function. To solve equation (28) we will rewrite equation (28) as + k*Z = 0 do? (30)

sin 8- + jG+1) sin² (@) = k² (27) sin e. and, (28) Z dộ? Show that equation (27) combined with equation (28) in a certain way gives us (26) (this means that equation (27) along with (28) is completely equivalent to equation(26)). [3] The solution for equation (27) requires power series analysis, which we will not go into; the solution is as follows: T (0) = T;x (0) = n;aP} (cos (0)), (29) here njk are constants that can be specified with integers j and k; P is a special function called Legendre function. To solve equation (28) we will rewrite equation (28) as + k*Z = 0 do? (30)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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d dT sin 0- + jG + 1)sin" (0)} + {a} (), = 0 (26) Z dộ? sin By the same logical reasoning we used to separate radial and angular part before, we will separate the left and right parts (in curly braces) from equation (26) to obtain the following dT sin 0- de + jlj + 1) sin? (8) = k2 sin 8- (27) OP and, 1 dZ (28) Z dộ? (G) Show that equation (27) combined with equation (28) in a certain way gives us (26) (this means that equation (27) along with (28) is completely equivalent to equation(26)). The solution for equation (27) requires power series analysis, which we will not go into; the solution is as follows: T (0) = T;+ (0) = n;APF (cos (0)), here n;4 are constants that can be specified with integers j and k; P is a special function called Legendre function. To solve equation (28) we (29) will rewrite equation (28) as + k²Z = 0 (30)
Transcribed Image Text:d dT sin 0- + jG + 1)sin" (0)} + {a} (), = 0 (26) Z dộ? sin By the same logical reasoning we used to separate radial and angular part before, we will separate the left and right parts (in curly braces) from equation (26) to obtain the following dT sin 0- de + jlj + 1) sin? (8) = k2 sin 8- (27) OP and, 1 dZ (28) Z dộ? (G) Show that equation (27) combined with equation (28) in a certain way gives us (26) (this means that equation (27) along with (28) is completely equivalent to equation(26)). The solution for equation (27) requires power series analysis, which we will not go into; the solution is as follows: T (0) = T;+ (0) = n;APF (cos (0)), here n;4 are constants that can be specified with integers j and k; P is a special function called Legendre function. To solve equation (28) we (29) will rewrite equation (28) as + k²Z = 0 (30)
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