Q7.3 - 12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

$\mathbf{\text{sin3t cos3t}}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{3}{s^2+36}$.

Definition of Laplace transform

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform. 
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by$\mathcal{L}\left\{f\left(t\right)\right\}$ or F(s).

Determine the Laplace transform for the given equation

Given that $\sin 3t\cos 3t$,

Find the Laplace transform of $\sin 3t\cos 3t$ using $\sin a\cos b=\frac{1}{2}(\sin a-b+\sin a+b)$ and $\mathcal{L}\left\{\sin bt\right\}=\frac{b}{s^2+b^2}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{\sin 3t\cos 3t\right\}& =& \mathcal{L}\left\{\frac{1}{2}\left[\sin \right(3t-3t)+\sin (3t+3t\left)\right]\right\}\\ & =& \frac{1}{2}\mathcal{L}\{\sin \left(0\right)+\sin 6t\}\\ & =& \frac{1}{2}\mathcal{L}\{0+\sin 6t\}\\ & =& \frac{1}{2}\mathcal{L}\left\{\sin 6t\right\}\end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{\sin 3t\cos 3t\right\}& =& \frac{1}{2}\left[\frac{6}{s^2+36}\right]\\ & =& \frac{3}{s^2+36}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{3}{s^2+36}$.