Question

In problems 47and 48use one of the inverses Laplace transform found in problems 31-54to solve the given initial-value problems.

${\mathit{y}}^{\mathbf{n}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{10}\mathit{c}\mathit{o}\mathit{s}\mathbf{5}\mathit{t}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The solved initial-value problem is below

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{\frac{10s}{\left(s^2+5^2\right)\left(s^2+2^2\right)}\right\}=\frac{10}{25-4}\left[cos2t-\cos 5t\right] \\ y\left(t\right)=\frac{10}{21}\left[\cos 2t-\cos 5t\right] \end{gathered}$$

Step-by-step solution

Step 1: definition of Laplace transform

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Calculate the Laplace transform for both sides of the given equation.

$\begin{array}{rcl}L\left\{y\text{'}\text{'}\right\}& =& s^{2}L\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)\left(*\right)\\ L\left\{10cos5t\right\}& =& L\left\{y\text{'}\text{'}+4y\right\}\end{array}$

$\begin{array}{rcl}& & \begin{array}{c}L\left\{10cos5t\right\}=L\left\{y\text{'}\text{'}+4y\right\}\\ =L\left\{y\text{'}\text{'}\right\}+4L\left\{y\right\}\\ ={\mathrm{s}}^{2}L\left\{y\right\}-sy\left(0\right)-y\text{'}\left(0\right)+4L\left\{y\right\}\end{array}\end{array}$

$=\left({\mathrm{s}}^2+4\right)L\left\{y\right\},\left(y\right.\left(0\right)=0,y\text{'}\left(0\right)=\left.0\right)$

$L\left\{y\right\}=Y\left(s\right)$and replace in the above equation.

role="math" localid="1663976388010" $\begin{array}{rcl}10L\left\{\cos 5t\right\}& =& \left(s^2+4\right)Y\left(s\right)\\ \frac{10s}{s^2+5^2}& =& \left(s^2+4\right)Y\left(s\right)\\ Y\left(s\right)& =& \frac{10s}{\left(s^2+25\right)\left(s^2+4\right)}\end{array}$

As a result, the required solution can be found by taking the inverse Laplace for the obtained function.

$y\left(t\right)=L^{-1}\left\{Y\left(s\right)\right\}=L^{-1}\left\{\frac{10s}{\left(s^2+25\right)\left(s^2+4\right)}\right\}$

Step2:determine the Laplace transformation.

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 given for t = 0 and assume the function meets certain conditions that will be stated later.

Now use the formula

$L^{-1}\left\{\frac{S}{\left(s^2+a^2\right)\left(s^2+b^2\right)}\right\}=\frac{1}{a^2-b^2}\left[cosbt-cosat\right]$

role="math" localid="1663976500536" $\begin{array}{rcl}y\left(t\right)& =& L^{-1}\left\{\frac{10s}{\left(s^2+5^2\right)\left(s^2+2^2\right)}\right\}=\frac{10}{25-4}\left[cos2t-\cos 5t\right]\\ & =& \frac{10}{21}\left[\cos 2t-\cos 5t\right]\end{array}$

Hence,

Laplace transformation,$y\left(t\right)=\frac{10}{21}\left[\cos 2t-\cos 5t\right]$