Q 19E - 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.]

$\mathbf{cosnt}\mathbf{ }\mathbf{sinmt}\mathbf{,}\mathbf{m}\mathbf{\ne }\mathbf{n}$

Step-by-Step Solution

Answer

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

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

Find the Laplace transform of $\cos nt\sin mt$ 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:

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

Simplify the equation as:

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

Further simplifying the equation as follows:

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

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