In the Previous Three Problems, the Solution of the Original $u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } }$

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

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

Correct Answer:

Verified

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

Access For Free