Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is
A) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
B) $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
C) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
D) $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
E) $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$

Question

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The Answer of Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is
A) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
B) $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
C) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$
D) $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$
E) $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$

Apply a Fourier Transform In xxx In the Previous Problem U(α,t)=F{u(x,t)}U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}U(α,t)=F{u(x,t)}

Apply a Fourier Transform In $x$ In the Previous Problem $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$