The Laplace transform of sin h (at) is

Laplace Transform of Hyperbolic Sine Function

The question asks for the Laplace transform of the function $\sinh (at)$. The hyperbolic sine function, $\sinh (x)$, is defined in terms of exponential functions. We can use this definition and the properties of the Laplace transform to find the answer.

Understanding Hyperbolic Sine

The hyperbolic sine function, denoted as $\sinh (x)$, is defined as:

$$\sinh (x)=\frac{e^x-e^{-x}}{2}$$

Therefore, the function we are considering is:

$$\sinh (at)=\frac{e^{at}-e^{-at}}{2}$$

Applying Laplace Transform Properties

The Laplace transform of a function $f(t)$, denoted by $\mathcal{L}\{f(t)\}$, is defined as an integral. Key properties, like linearity, are very useful. The linearity property states that $\mathcal{L}\{c_1{f}_1(t)+c_2{f}_2(t)\}=c_1\mathcal{L}\{f_1(t)\}+c_2\mathcal{L}\{f_2(t)\}$.

We need to find $\mathcal{L}\{\sinh (at)\}$. Using the definition:

$$\mathcal{L}\{\sinh (at)\}=\mathcal{L}\left\{\frac{e^{at}-e^{-at}}{2}\right\}$$

Using the linearity property, we can take the constant $\frac{1}{2}$ out and apply the transform to the difference of the exponential functions:

$$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}\left(\mathcal{L}\{e^{at}\}-\mathcal{L}\{e^{-at}\}\right)$$

Standard Laplace Transforms

We use the known standard Laplace transform pairs:

- The Laplace transform of $e^{bt}$ is $\frac{1}{s-b}$.

Applying this rule to our terms:

- $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$
- $\mathcal{L}\{e^{-at}\}=\frac{1}{s-(-a)}=\frac{1}{s+a}$

Deriving the Final Formula

Now, substitute these results back into our equation:

$$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}\left(\frac{1}{s-a}-\frac{1}{s+a}\right)$$

To simplify the expression inside the parentheses, find a common denominator:

$$\frac{1}{s-a}-\frac{1}{s+a}=\frac{(s+a)-(s-a)}{(s-a)(s+a)}$$

Simplify the numerator:

$$(s+a)-(s-a)=s+a-s+a=2a$$

The denominator is a difference of squares:

$$(s-a)(s+a)=s^2-a^2$$

So the expression becomes:

$$\frac{2a}{s^2-a^2}$$

Substitute this back into the main equation:

$$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}\left(\frac{2a}{s^2-a^2}\right)$$

Simplify by canceling the factor of 2:

$$\mathcal{L}\{\sinh (at)\}=\frac{a}{s^2-a^2}$$

Conclusion

Comparing our result with the given options, we find that the correct Laplace transform for $\sinh (at)$ is $\frac{a}{s^2-a^2}$.