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In the Three Previous Problems, the Solution for the Temperature u(x,t)u ( x , t )

Question 2

Multiple Choice

In the three previous problems, the solution for the temperature u(x,t) u ( x , t ) is


A) u=u00[sin(αx) (1ekα2t) /α]dαu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha
B) u=u00[sin(αx) (1ekα2t) /α]dα/πu = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi
C) u=2u00[sin(αx) (1ekα2t) /α]dαu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha
D) u=2u00[sin(αx) (1ekα2t) /α]dα/πu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi
E) u=2u00[sin(αx) (1ekα2t) /α]dα/πu = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi

Correct Answer:

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