Question

In Problems 15 and 16 use a graphing utility to graph the indicated

solution.

y (t) of Problem $\mathbf{0}\le \mathbf{t}<\mathbf{2}\pi$

Short answer

Graph lies between 0 and 6 as the t value lies between $0\le t<2\pi$.

Step-by-step solution

Laplace transformation of a linear differential equation with constant coefficient

1. The Laplace transformation of a linear differential equation with constant coefficient is express a.${\mathbf{a}}_n\mathbf{L}\left\{\frac{d^{n}y}{{\mathrm{dt}}^n}\right\}\mathbf{=}{\mathbf{a}}_n\left[{\mathbf{s}}^n\mathbf{Y}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{-}{\mathbf{s}}^{(n-1)}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{-}\mathbf{L}\mathbf{-}{\mathbf{y}}^{(n-1)}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\right]$.

2. The formula for the inverse Laplace transform is as follows,

3. The formula for the Laplace transform of $\cos kt$ is as follows,

$\mathbf{L}\mathbf{\left\{}\mathbf{coskt}\mathbf{\right\}}\mathbf{=}\frac{s}{{\mathbf{s}}^2\mathbf{+}{\mathbf{k}}^2}$

Apply the Laplace transform of the given linear differential equation

Use the above definition to write the Laplace transform of the given linear differential equation.

${\mathrm{s}}^2\mathrm{L}\left\{\mathrm{y}\right\}-\mathrm{sy}\left(0\right)-{\mathrm{y}}^{\text{'}}\left(0\right)+16\mathrm{L}\left\{\mathrm{y}\right\}=\mathrm{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}$

Use the above definition to write the function$f\left(t\right)$ as follows,

$f\left(t\right)=\cos 4t-\cos 4tU(t-\pi )$

Put the value $y\left(0\right)=0$and $y^{\text{'}}\left(0\right)=1$in above expression.

Above expression is simplified as,

From the above expression,

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{({\mathrm{s}}^2+16)}+\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}-\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}----\left(1\right)$

Evaluate the coefficients of s2

Above expression is simplified as follows,

$\begin{array}{l}\mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{({\mathrm{s}}^2+16)}+\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}-\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}\\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{4}\frac{4}{({\mathrm{s}}^2+16)}+\frac{1}{8}\frac{8\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}-\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}\end{array}$

From the above expression,

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{4}\frac{4}{({\mathrm{s}}^2+16)}+\frac{1}{8}\frac{8\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}-\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}---\left(2\right)$

Apply inverse Laplace transform

Take inverse Laplace on both sides of equation (2),

$\begin{array}{c}\mathrm{y}={\mathrm{L}}^{-1}\left\{\frac{1}{4}\frac{4}{({\mathrm{s}}^2+16)}+\frac{1}{8}\frac{8\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}-\frac{\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}\right\}\\ =\frac{1}{4}{\mathrm{L}}^{-1}\left\{\frac{4}{{\mathrm{s}}^2+16}\right\}+\frac{1}{8}{\mathrm{L}}^{-1}\left\{\frac{8\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}\right\}-\frac{1}{8}{\mathrm{L}}^{-1}\left\{\frac{8\mathrm{s}}{{({\mathrm{s}}^2+16)}^2}{\mathrm{e}}^{-\mathrm{\pi s}}\right\}\\ =\frac{1}{4}\sin 4\mathrm{t}+\frac{1}{8}\mathrm{tsin}4\mathrm{t}-\frac{1}{8}(\mathrm{t}-\mathrm{\pi })\sin 4(\mathrm{t}-\mathrm{\pi })\mathrm{U}(\mathrm{t}-\mathrm{\pi })\end{array}$

Therefore, the solution of given initial value problem is

_{$\mathrm{y}=\frac{1}{4}\sin 4\mathrm{t}+\frac{1}{8}\mathrm{tsin}4\mathrm{t}-\frac{1}{8}(\mathrm{t}-\mathrm{\pi })\sin 4(\mathrm{t}-\mathrm{\pi })\mathrm{U}(\mathrm{t}-\mathrm{\pi })$}

Plot the graph for the function

The graph lies between 0 and 6 as the t value lies between $0\le \mathrm{t}<2\mathrm{\pi }$.