Find the inverse Laplace transform of each of the following expressions: (a) \(\frac{4}{s^{2}+4 s+5}\) (b) \(\frac{s+5}{s^{2}+10 s+29}\) (c) \(\frac{2 s-3}{s^{2}+6 s+10}\) (d) \(\frac{s^{2}+6 s-4}{\left(s^{2}+4\right)^{2}}\) (e) \(\frac{1}{4 s^{2}+4 s+1}\)

Let's recognize \(\frac{4}{s^{2}+4 s+5}\) as the Laplace transform of some function. Note that the denominator is a quadratic, and we need to complete the square to determine if this Laplace Transform is known. By completing the square, the quadratic can be re-written as \((s+2)^{2}+1\). This expression now looks like the Laplace Transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\).

In our case, \(a=2\) and \(b=1\). Applying the inverse Laplace transform, we have \(\mathcal{L}^{-1}\left\{\frac{4}{(s+2)^{2}+1}\right\} = 4 e^{-2t}\sin{(1t)} = 4e^{-2t}\sin{t}\). So, the inverse Laplace transform of \(\frac{4}{s^{2}+4 s+5}\) is \(4e^{-2t}\sin{t}\). # (b) #

Let's recognize \(\frac{s+5}{s^{2}+10 s+29}\) as the Laplace Transform of some function. By completing the square for the quadratic in the denominator, we can rewrite it as \((s+5)^{2}+4\). This expression now looks like the Laplace Transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\).

In our case, \(a=5\) and \(b=2\). Applying the inverse Laplace transform, we have \(\mathcal{L}^{-1}\left\{\frac{s+5}{(s+5)^{2}+4}\right\} = e^{-5t}\cos{(2t)}\). So, the inverse Laplace transform of \(\frac{s+5}{s^{2}+10 s+29}\) is \(e^{-5t}\cos{2t}\). # (c) #

Let's recognize \(\frac{2s-3}{s^{2}+6 s+10}\) as the Laplace Transform of some function. By completing the square for the quadratic in the denominator, we can rewrite it as \((s+3)^{2}+1\). Now, as the Laplace Transform of the first-order damped sine function, this expression looks like either \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) or \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\). In order to align with one of these forms, you need to rewrite \(\frac{2s-3}{(s+3)^{2}+1}\) as the sum/division of two terms, each of which corresponds to one of these known rules.

Rewrite the numerator in terms of \((s+3)\). In other words, we rewrite \((2s-3)\) as \(q(s+3)+r\). The variables \(q\) and \(r\) are determined by solving the linear equation system: \(2s-3=q(s+3)+r\). In our case, we have \(q=2\) and \(r=-9\).

Now, we can rewrite our expression as \(\frac{2(s+3)-9}{(s+3)^{2}+1}\). This can be further separated into two parts: \(\frac{2(s+3)}{(s+3)^{2}+1}-\frac{9}{(s+3)^{2}+1}\). The first part matches the inverse Laplace transform \(\mathcal{L}\{e^{-at}\cos{(bt)}\}=\frac{s+a}{(s+a)^{2}+b^{2}}\) with \(a=3\) and \(b=1\). The second part matches the inverse Laplace transform \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\) with \(a=3\) and \(b=1\). Therefore, we have the inverse Laplace transform for expression (c) as \(2e^{-3t}\cos{t}-9e^{-3t}\sin{t}\). # (d) #

Let's recognize \(\frac{s^{2}+6s-4}{\left(s^{2}+4\right)^{2}}\) as the Laplace Transform of some function. First, rewrite the numerator by completing the square. This results in the expression: \(\frac{(s+3)^{2}-13}{\left(s^{2}+4\right)^{2}}\).

From the table of Laplace transforms, we see that this looks like the Laplace transform pair \(\mathcal{L}\{t e^{at} \sin{(bt)}\} =\frac{2a s+b}{\left(s^2+b^2\right)^2}\) or \(\mathcal{L}\{t e^{at} \cos{(bt)}\} =\frac{s^2 - b^2}{\left(s^2+b^2\right)^2}\). We choose the second pair, because the numerator doesn't contain \(s\). In our case, we have \(a=0\) and \(b=2\). Because of the difference in the coefficients, we will find \(\mathcal{L}\{t \cos{2t}\}\) first and then adjust the coefficients. The inverse Laplace transform of \(\frac{s^2 - 4}{\left(s^2+4\right)^{2}}\) is \(t \cos{2t}\). Now, we need to multiply \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{s^2-4}\) in the time domain to get the inverse Laplace transform of the expression given.

In the time domain, multiplying \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{s^2-4}\) is the same as multiplying \(t\cos{2t}\) by \(\frac{(s+3)^{2}-13}{-8}\). Therefore, the inverse Laplace transform of \(\frac{s^{2}+6s-4}{\left(s^{2}+4\right)^{2}}\) is \(-\frac{1}{8}(3+t)^{2}t\cos{2t}+\frac{13}{8}t\cos{2t}\). # (e) #

Let's recognize \(\frac{1}{4 s^{2}+4 s+1}\) as the Laplace Transform of some function. We identify the denominator as the square of a first-order damped sine function, \(\left(\frac{s+1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\).

From the table of Laplace transforms, we look for an entry with a denominator in the same form. This pair looks like the Laplace transform of the first-order damped sine function, which is given by \(\mathcal{L}\{e^{-at}\sin{(bt)}\}=\frac{b}{(s+a)^{2}+b^{2}}\). So, let's apply the inverse Laplace transform of this form to our expression with \(a=-1\) and \(b=\frac{1}{2}\). Therefore, we have the inverse Laplace transform of \(\frac{1}{4 s^{2}+4 s+1}\) is \(2e^{t}\sin{\frac{t}{2}}\).