Inverse laplace transform of 𝐿−11𝑠2+𝑎2L −1 s 2 +a 2 1​ =a)1𝑎𝑐𝑜𝑠𝑎𝑡a1​ cosatb)1𝑎𝑠𝑖𝑛𝑎𝑡a1​ sinatc)1𝑎𝑠𝑖𝑛𝑡a1​ sintd)1𝑎𝑐𝑜𝑠𝑡a1​ cost

1. Break Down the Problem

We need to find the inverse Laplace transform of the expression $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\}$. The goal is to identify which of the given options matches this inverse transform.

2. Relevant Concepts

The formula for the inverse Laplace transform of $\frac{1}{s^2 + a^2}$ is given by: $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{1}{a} \sin(at)$ This is a well-known result in the context of Laplace transforms.

3. Analysis and Detail

Let us apply the known formula directly:

1. The original function in the Laplace domain is $\frac{1}{s^2 + a^2}$.
2. According to the inverse Laplace transform property, we know that this corresponds to $\frac{1}{a} \sin(at)$ in the time domain.

4. Verify and Summarize

To verify, we substitute back:

• The function $\frac{1}{s^2 + a^2}$ should yield a correct form.
• This agrees with our derived form, confirming the correctness of our approach.

Final Answer

The inverse Laplace transform of $\frac{1}{s^2 + a^2}$ is: $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{1}{a} \sin(at)$ Thus, the correct option is: (b) $\frac{1}{a} \sin(at)$.