Discussion

The solution \( y \) of \( y'' - y = 0 \) by Laplace transform, when \( y(0) = 12 \) and \( y'(0) = 0 \), is

Answer: A

Transforming \( y'' - y = 0 \) with \( y(0) = 12 \) and \( y'(0) = 0 \):

\[ s^{2}Y - s \cdot 12 - 0 - Y = 0 \]

\[ Y(s^{2} - 1) = 12s \quad\Longrightarrow\quad Y = \frac{12s}{s^{2}-1} \]

Since \( \mathcal{L}\{\cosh t\} = \dfrac{s}{s^{2}-1} \):

\[ y = 12\cosh t \]

The hyperbolic cosine appears because the denominator is \( s^{2} - 1 \); an ordinary cosine would need \( s^{2} + 1 \).

No comment is present. Be the first to comment.

Loading…