Question

(a) Assume that Theorem 7.3.1 holds when the symbol a is replaced by ki, where k is a real number and ${\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$. Show that $\mathbf{L}\left\{{\mathrm{te}}^{\mathrm{kti}}\right\}$ can be used to deduce

$$\begin{gathered} \mathcal{L}\mathbf{\left\{}\mathbf{tcoskt}\mathbf{\right\}}\mathbf{=}\frac{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}{{\left(s^2+k^2\right)}^{\mathbf{2}}} \\ \mathcal{L}\mathbf{\left\{}\mathbf{tsinkt}\mathbf{\right\}}\mathbf{=}\frac{\mathbf{2}\mathbf{ks}}{{\left(s^2+k^2\right)}^{\mathbf{2}}} \end{gathered}$$

(b) Now use the Laplace transform to solve the initial value problem

${\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{cos\omega t}\mathbf{,}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

(a)The given equation is proved.

$\mathcal{L}\left\{\mathrm{tcoskt}\right\}=\frac{{\mathrm{s}}^2-{\mathrm{k}}^2}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}\text{\hspace{1em}and\hspace{1em}}\mathcal{L}\left\{\mathrm{tsinkt}\right\}=\frac{2\mathrm{ks}}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}$

(b)The initial value equation is $\mathrm{x}\left(\mathrm{t}\right)=\frac{1}{2\mathrm{\omega }}\mathrm{tsin\omega t}$.

Step-by-step solution

Definition of Euler’s theorem,

According to Euler's Theorem, if $\gcd (\mathrm{a},\mathrm{n})=1$, then$\mathrm{a}\left(\mathrm{n}\right)=1.\left(\mathrm{mod}\mathrm{n}\right).\left(\mathrm{n}\right)$ is Euler's totient function, which counts the number of integers in the ranges$1,2,...,\mathrm{n}-1$ that are relatively prime to $n$.This theorem is just Fermat's little theorem when n is a prime.

Euler’s formula

(a)

If $\mathrm{F}\left(\mathrm{s}\right)=\mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}$, then$\mathcal{L}\left\{{\mathrm{t}}^{\mathrm{n}}\mathrm{f}\left(\mathrm{t}\right)\right\}={(-1)}^{\mathrm{n}}\frac{{\text{d}}^{\mathrm{n}}}{\text{d}{\mathrm{s}}^{\mathrm{n}}}\mathrm{F}\left(\mathrm{s}\right)$

Let $n=1$, then

Euler's formula

Based on$\left(1\right)$ and $\left(3\right)$, we have

$\mathcal{L}\left\{\mathrm{tcoskt}\right\}+\mathrm{i}\mathcal{L}\left\{\mathrm{tsinkt}\right\}=\frac{1}{{(\mathrm{s}-\mathrm{ki})}^2}$

$=\frac{{(\mathrm{s}+\mathrm{ki})}^2}{{(\mathrm{s}-\mathrm{ki})}^2{(\mathrm{s}+\mathrm{ki})}^2}$

$=\frac{{\mathrm{s}}^2+2\mathrm{ksi}-{\mathrm{k}}^2}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}$

$=\frac{{\mathrm{s}}^2-{\mathrm{k}}^2}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}+\mathrm{i}\frac{2\mathrm{ks}}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}$

Comparing the real and imaginary parts, we get

$\mathcal{L}\left\{\mathrm{tcoskt}\right\}=\frac{{\mathrm{s}}^2-{\mathrm{k}}^2}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}\text{\hspace{1em}and\hspace{1em}}\mathcal{L}\left\{\mathrm{tsinkt}\right\}=\frac{2\mathrm{ks}}{{({\mathrm{s}}^2+{\mathrm{k}}^2)}^2}$

Hence, its proved..

Solve the initial value problem.

(b)

Given that

$\{\begin{array}{c}\mathcal{L}\left\{{\mathrm{x}}^{\text{'}\text{'}}\right\}={\mathrm{s}}^2\mathcal{L}\left\{\mathrm{x}\right\}-\mathrm{sx}\left(0\right)-{\mathrm{x}}^{\text{'}}\left(0\right)\\ \mathrm{x}\left(0\right)=0,{\mathrm{x}}^{\text{'}}\left(0\right)=0\end{array}$

Take Laplace transform of both side of the given equation.

$\begin{array}{l}ℒ\left\{\cos \omega t\right\}=ℒ\left\{x^{\text{'}\text{'}}\right\}+{\omega }^2ℒ\left\{x\right\}\\ ℒ\left\{\cos \omega t\right\}=s^2ℒ\left\{x\right\}+{\omega }^2ℒ\left\{x\right\}\\ \frac{s}{s^2+{\omega }^2}=(s^2+{\omega }^2)ℒ\left\{x\right\}\left(\mathcal{L}\left\{\mathrm{coskt}\right\}=\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right)\end{array}$

Denote$\mathcal{L}\left\{\mathrm{x}\right\}=\mathrm{X}\left(\mathrm{s}\right)$and replace in the above equation.

$\mathrm{X}\left(\mathrm{s}\right)=\frac{\mathrm{s}}{{({\mathrm{s}}^2+{\mathrm{\omega }}^2)}^2}$

Apply inverse Laplace transform to each side.

Therefore, the initial value equation is $\mathrm{x}\left(\mathrm{t}\right)=\frac{1}{2\mathrm{\omega }}\mathrm{tsin\omega t}$.