In the Previous Three Problems, the Solution of the Original $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$

Multiple Choice

In the previous three problems, the solution of the original problem is

A) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$ , where $c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )$
B) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )$ , where $c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
C) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh ( n \pi ( y - H ) / L )$ , where $c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
D) $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)$ , where $c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )$
E) $u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \sinh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)$ , where $c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )$