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The Solution of the Previous Three Problems Is ∑n=1∞cnXn(x)e−λnkt\sum _ { n = 1 } ^ { \infty } c _ { n } X _ { n } ( x ) e ^ { - \lambda _ { n } k t }

Question 28

Multiple Choice

The solution of the previous three problems is ∑n=1∞cnXn(x) e−λnkt\sum _ { n = 1 } ^ { \infty } c _ { n } X _ { n } ( x ) e ^ { - \lambda _ { n } k t } , where XnX _ { n } and λn\lambda _ { n } are given in the previous problem and


A) cn=∫0Lf(x) dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) d x
B) cn=∫0Lf(x) Xn(x) dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
C) cn=∫0Lf(x) Xn(x) dx/∫0LXn(x) dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ( x ) d x
D) cn=∫0LXn2(x) dx/∫0Lf(x) Xn(x) dxc _ { n } = \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x / \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
E) cn=∫0Lf(x) Xn(x) dx/∫0LXn2(x) dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x

Correct Answer:

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