Question

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}$$\begin{array}{l}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\mathbf{cos}\mathbf{4}\mathbf{t}\right\}\mathbf{,}\mathbf{0}\mathbf{}\mathit{\epsilon }\mathbf{}\mathbf{t}\mathbf{<}\mathbf{\pi }\\ \left\{\mathbf{0}\right\}\mathbf{,}{\mathbf{t}}^{\mathbf{3}}\mathbf{\pi }\end{array}$

Short answer

Solution of given initial value problem is

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

Step-by-step solution

Laplace transformation of a linear differential equation with constant coefficient

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

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

$\begin{array}{l}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{\mathbf{k}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}\right\}\mathbf{=}\mathbf{sinkt}\\ {\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{\mathbf{s}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}\right\}\mathbf{=}\mathbf{coskt}\\ {\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{\mathbf{2}\mathbf{ks}}{{\left({\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}\right)}^{\mathbf{2}}}\right\}\mathbf{=}\mathbf{tsinkt}\end{array}$

3. The formula for the Laplace transform of is as follows,

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

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,

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

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

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

Above expression is simplified as,

From the above expression,

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

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),

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

Therefore, the solution of given initial value problem is

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