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

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

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

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free