Q7.3 - 15E - 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.]

${\mathbf{cos}}^{\mathbf{3}}\mathbf{t}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{s^3+7s}{\left(s^2+9\right)\left(s^2+1\right)}$.

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 ${\cos }^{3}t$,

Find the Laplace transform of ${\cos }^{3}t$ using $co{s}^{3}a=\frac{1}{4}(cos3a+3cosa)$, $\mathcal{L}\left\{af\right(x)\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, $\mathcal{L}\left\{cosbt\right\}=\frac{s}{s^2+b^2}$ and $\frac{a}{b}\pm \frac{c}{d}=\frac{da\pm bc}{bd}$ as:

$\begin{array}{rcl}\mathcal{L}\left\{{\cos }^{3}t\right\}& =& \mathcal{L}\left\{\frac{1}{4}(\cos 3t+3\cos t)\right\}\\ & =& \frac{1}{4}\mathcal{L}\{\cos 3t+3\cos t\}\\ & =& \frac{1}{4}[\mathcal{L}\{\cos 3t\}+3\mathcal{L}\{\cos t\left\}\right]\\ & =& \frac{1}{4}\left[\frac{s}{s^2+3^2}+3\times \frac{s}{s^2+1^2}\right]\end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{{\cos }^{3}t\right\}& =& \frac{1}{4}\left[\frac{s}{s^2+9}+\frac{3s}{s^2+1}\right]\\ & =& \frac{1}{4}\left[\frac{s\left(s^2+1\right)+3s\left(s^2+9\right)}{\left(s^2+9\right)\left(s^2+1\right)}\right]\\ & =& \frac{1}{4}\left[\frac{s^3+s+3s^3+27s}{\left(s^2+9\right)\left(s^2+1\right)}\right]\\ & =& \frac{1}{4}\left[\frac{{\overbrace{4s^3+28s}}^{4\text{ Common }}}{\left(s^2+9\right)\left(s^2+1\right)}\right]\end{array}$

Further simplifying the equation as follows:

$\begin{array}{rcl}\mathcal{L}\left\{{\cos }^{3}t\right\}& =& 4\left[\frac{4\left(s^3+7s\right)}{\left(s^2+9\right)\left(s^2+1\right)}\right]\\ & =& \frac{s^3+7s}{\left(s^2+9\right)\left(s^2+1\right)}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{s^3+7s}{\left(s^2+9\right)\left(s^2+1\right)}$.