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In the Previous Two Problem, the Solution For u(x,t)u ( x , t )

Question 19

Multiple Choice

In the previous two problem, the solution for u(x,t) u ( x , t ) is


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

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