In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is
A) $U = \sinh ( s x ) + \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$
B) $U = \cosh ( s x ) + \frac { s } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$
C) $U = \frac { 1 } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$
D) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$
E) $U = \frac { 1 } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$

Question

Quizplus · Accepted Answer

The correct solution for $U(x, s)$ when $f(x) = \sin(\pi x)$ in the context of Laplace transforms involves finding a function that, when transformed back, gives the original function $f(x)$. The term $\frac{s}{s^2 + \pi^2}$ is a common Laplace transform pair for $\sin(\pi x)$, indicating that option D correctly represents the solution $U(x, s)$ as it incorporates the Laplace transform of $\sin(\pi x)$ correctly.

In the Previous Problem, Assume That f(x)=sin⁡(πx)f ( x ) = \sin ( \pi x )f(x)=sin(πx)

In the Previous Problem, Assume That $f ( x ) = \sin ( \pi x )$