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In the Previous Four Problems, the Infinite Series Solution Is u=n1cnJ0(nπr/3)cos(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )

Question 28

Multiple Choice

In the previous four problems, the infinite series solution is (for certain values of the constants)


A) u=n1cnJ0(nπr/3) cos(nπz/3) u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )
B) u=n1cnJ0(n2π2r/9) cos(nπz/3) u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \cos ( n \pi z / 3 )
C) u=n1cnJ0(n2π2r/9) sin(nπz/3) u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \sin ( n \pi z / 3 )
D) u=n1cnJ0(znr/2) sinh(zn(z3) /2) u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \sinh \left( z _ { n } ( z - 3 ) / 2 \right)
E) u=n=1cnJ0(znr) cosh(zn(z3) ) u = \sum _ { n = 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r \right) \cosh \left( z _ { n } ( z - 3 ) \right)

Correct Answer:

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