In the previous problem, apply the Laplace transform. The resulting equation for $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$ is
A) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = - s f ( x )$
B) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = f ( x )$
C) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - f ( x )$
D) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = s f ( x )$
E) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - s f ( x )$

Question

Quizplus · Accepted Answer

The Laplace transform of a differential equation typically involves a transformation of the time derivative into a multiplication by $s$ in the Laplace domain, and the initial conditions are incorporated into the transformed equation. The correct equation, $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - s f ( x )$, correctly reflects the transformation of a second-order time-dependent differential equation into the Laplace domain, where $s$ is the Laplace variable, and incorporates the initial condition through the term $- s f ( x )$.

In the Previous Problem, Apply the Laplace Transform U(x,s)=L{u(x,t)}U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}U(x,s)=L{u(x,t)}

In the Previous Problem, Apply the Laplace Transform $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$