Q7.3 - 17E - 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.]

$\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{ }\mathbf{sin}\mathbf{5}\mathbf{t}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{20s}{\left(s^2+9\right)\left(s^2+49\right)}$.

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 $\sin 2t\sin 5t$,

Find the Laplace transform of $\sin 2t\sin 5t$ using $sinasinb=\frac{1}{2}\left[cos\right(a-b)-cos(a+b\left)\right]$, $cos(-x)=cosx$,  $\mathcal{L}\left\{af\right(x)\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, $\mathcal{L}\left\{cosbt\right\}=\frac{s}{s^2+b^2}$ and  $\frac{a}{c}\pm \frac{b}{d}=\frac{da\pm cb}{cd}$as:

$\begin{array}{rcl}\mathcal{L}\left\{\sin 2t\sin 5t\right\}& =& \mathcal{L}\left\{\frac{1}{2}\left[\cos \right(2t-5t)-\cos (2t+5t\left)\right]\right\}\\ & =& \frac{1}{2}[\mathcal{L}\{\cos 3t\}-\mathcal{L}\{\cos 7t\left\}\right]\\ & =& \frac{1}{2}\left[\frac{s}{s^2+9}-\frac{s}{s^2+49}\right]\\ & =& \frac{1}{2}\left[\frac{\left(s^2+49\right)\times s-\left(s^2+9\right)\times s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{\sin 2t\sin 5t\right\}& =& \frac{1}{2}\left[\frac{s^3+49s-s^3-9s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\\ & =& \frac{1}{2}\left[\frac{40s}{\left(s^2+9\right)\left(s^2+49\right)}\right]\\ & =& \frac{20s}{\left(s^2+9\right)\left(s^2+49\right)}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{20s}{\left(s^2+9\right)\left(s^2+49\right)}$.