Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform. Complete parts a and b below. \( \mathscr{L}\left\{e^{9 t} \sin 7 t-t^{5}+e^{4 t}\right\} \) Click the icon to view the Laplace transform table. a. Determine the formula for the Laplace transform. \( \mathcal{L}^{9}\left\{e^{9 t} \sin 7 t-t^{5}+e^{4 t}\right\}=\square \) (Type an expression using s as the variable.) b. What is the restriction on s? \( s>\square \) (Type an integer or a fraction.)

To find the Laplace transform of \( e^{9 t} \sin 7 t - t^{5} + e^{4 t} \), we can use the linearity of the Laplace transform, which allows us to compute the transform of each term independently. a. According to the formulae from the Laplace transform table: - The Laplace transform of \( e^{at} \sin bt \) is given by \( \frac{b}{(s-a)^2 + b^2} \). - The Laplace transform of \( t^{n} \) is \( \frac{n!}{s^{n+1}} \). - The Laplace transform of \( e^{bt} \) is \( \frac{1}{s-b} \). Applying these rules: 1. For \( e^{9t} \sin(7t) \): \[ \mathscr{L}\{ e^{9t} \sin(7t) \} = \frac{7}{(s-9)^2 + 7^2} = \frac{7}{(s-9)^2 + 49} \] 2. For \( -t^5 \): \[ \mathscr{L}\{ -t^5 \} = -\frac{5!}{s^{6}} = -\frac{120}{s^6} \] 3. For \( e^{4t} \): \[ \mathscr{L}\{ e^{4t} \} = \frac{1}{s-4} \] Combining these transforms, we have: \[ \mathscr{L}\left\{e^{9 t} \sin 7 t - t^{5} + e^{4 t}\right\} = \frac{7}{(s-9)^2 + 49} - \frac{120}{s^6} + \frac{1}{s-4} \] So, the answer is: \[ \mathcal{L}\left\{e^{9 t} \sin 7 t - t^{5} + e^{4 t}\right\} = \frac{7}{(s-9)^2 + 49} - \frac{120}{s^6} + \frac{1}{s - 4} \] b. The restrictions on \( s \) come from the exponential terms and the polynomial term: - For \( e^{9t} \sin(7t} \), we need \( s > 9 \). - For the term \( e^{4t} \), we need \( s > 4 \). - The polynomial term \( -t^{5} \) does not impose any additional restrictions. Thus, the most stringent condition is \( s > 9 \): \[ s > 9 \]