Q7.3 - 18E - 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.]

$\mathbf{cos}\mathbf{ }\mathbf{nt}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{mt}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{ }\mathbf{\ne }\mathbf{n}$

Step-by-Step Solution

Answer

The Laplace transform of for $m\ne n$is $\frac{\mathrm{s}\left({\mathrm{s}}^2+{\mathrm{n}}^2+{\mathrm{m}}^2\right)}{\left({\mathrm{s}}^2+{(\mathrm{n}-\mathrm{m})}^2\right)\left({\mathrm{s}}^2+{(\mathrm{n}+\mathrm{m})}^2\right)}$.

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 $\cos nt\cos mt, m\ne n$,

Find the Laplace transform of $\cos nt\cos mt$ for $m\ne n$ using $\cos a\cos b=\frac{1}{2}\left[\cos \right(a-b)+\cos (a+b\left)\right]$, $\mathcal{L}\left\{af\right(x)\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, $\mathcal{L}\left\{\cos bt\right\}=\frac{s}{s^2+b^2}$, $\frac{a}{c}\pm \frac{b}{d}=\frac{da\pm cb}{cd}$ and $(a\pm b{)}^2=a^2\pm 2ab+b^2$ as:

$\begin{array}{rcl}\mathcal{L}\left\{\cos nt\cos mt\right\}& =& \mathcal{L}\left\{\frac{1}{2}\left[\cos \right(nt-mt)+\cos (nt+mt\left)\right]\right\}\\ & =& \frac{1}{2}\left[\mathcal{L}\left\{\cos \right(n-m\left)t\right\}+\mathcal{L}\left\{\cos \right(n+m\left)t\right\}\right]\\ & =& \frac{1}{2}\left[\frac{s}{s^2+{(n-m)}^2}+\frac{s}{s^2+{(n+m)}^2}\right]\\ & =& \frac{1}{2}\left[\frac{\left(s^2+{(n+m)}^2\right)·s+\left(s^2+{(n-m)}^2\right)·s}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\right]\end{array}$

Simplify the equation as:

$$\begin{gathered} \mathcal{L}\left\{\cos nt\cos mt\right\}=\frac{1}{2}\left[\frac{s^3+s{(n+m)}^2+s^3+s{(n-m)}^2}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\right] \\ =\frac{1}{2}\left[\frac{\stackrel{s Common}{\stackrel{⏜}{2s^3+s\left(n^2+2nm+m^2\right)+s\left(n^2-2nm+m^2\right)}}}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\right] \\ =\frac{1}{2}\left[\frac{s\left(2s^2+n^2+2nm+m^2+n^2-2nm+m^2\right)}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\right] \\ =\frac{1}{2}\left[\frac{\stackrel{2 Common}{\stackrel{⏜}{s\left(2s^2+2n^2+2m^2\right)}}}{{\displaystyle \left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}}\right] \end{gathered}$$

Further simplifying the equation as follows:

$\begin{array}{rcl}\mathcal{L}\left\{\cos nt\cos mt\right\}& =& \frac{1}{2}\left[\frac{2s\left(s^2+n^2+m^2\right)}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\right]\\ & =& \frac{s\left(s^2+n^2+m^2\right)}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}\end{array}$

Hence, the Laplace transform is $\frac{s\left(s^2+n^2+m^2\right)}{\left(s^2+{(n-m)}^2\right)\left(s^2+{(n+m)}^2\right)}$.