Solved

In the Previous Two Problems, the Solution For u(x,y)u ( x , y )

Question 3

Multiple Choice

In the previous two problems, the solution for u(x,y) u ( x , y ) is


A) u=∑n=1∞cncos⁡(nπx) sinh⁡(nπx) , where cn=∫02f(x) cos⁡(nπx) dx/sinh⁡(2nπ) u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi x ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
B) u=∑n=1∞cnsin⁡(nπx) sinh⁡(nπy) , where cn=∫02f(x) cos⁡(nπx) dx/sinh⁡(2nπ) u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \sinh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
C) u=c0y+∑n=1∞cncos⁡(nπx) sinh⁡(nπy) u = c _ { 0 } y + \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \sinh ( n \pi y ) , where cn=∫02f(x) dx/4c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) d x / 4 and
cn=∫02f(x) cos⁡(nπx) dx/sinh⁡(2nπ) c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
D) u=∑n=1∞cncos⁡(nπx) cosh⁡(nπy) , where cn=∫02f(x) cos⁡(nπx) dx/sinh⁡(2nπ) u = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )
E) u=∑n=1∞cnsin⁡(nπx) cosh⁡(nπy) , where cn=∫02f(x) cos⁡(nπx) dx/sinh⁡(2nπ) u = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x ) \cosh ( n \pi y ) , \text { where } c _ { n } = \int _ { 0 } ^ { 2 } f ( x ) \cos ( n \pi x ) d x / \sinh ( 2 n \pi )

Correct Answer:

verifed

Verified

Related Questions