Chapter 7.6, Problem 1P

Program Plan Intro

Program Description: Purpose of the problem is to solve the initial value problem $x^{″}+4x=\delta \left(t\right);x\left(0\right)=x^{\prime }\left(0\right)=0$ and graph the solution.

Explanation of Solution

Given information:

The differential equation is written as,

  $x^{″}+4x=\delta \left(t\right)$

The initial values for differential equation is written as,

  $\begin{array}{c}x\left(0\right)=0\\ x^{\prime }\left(0\right)=0\end{array}$

Calculation:

Take the Laplace transform of the given differential equation and solve as shown below.

  $\begin{array}{c}L\left\{x^{″}\left(t\right)+4x\left(t\right)\right\}=L\left\{\delta \left(t\right)\right\}\\ L\left\{x^{″}\left(t\right)\right\}+4L\left\{x\left(t\right)\right\}=e^{\left(0\right)t}\end{array}$

Further, simplify the expression as,

  $\begin{array}{l}{s}^{2}L\left\{x\left(t\right)\right\}-sx\left(0\right)-x^{\prime }\left(0\right)+4L\left\{x\left(t\right)\right\}=1\\ s^{2}L\left\{x\left(t\right)\right\}-s\left(0\right)-\left(0\right)+4L\left\{x\left(t\right)\right\}=1\\ \left(s^2+4\right)L\left\{x\left(t\right)\right\}=1\\ L\left\{x\left(t\right)\right\}=\frac{1}{s^2+4}\end{array}$

Further, simplify the expression as,

  $\begin{array}{c}x\left(t\right)=L^{-1}\left\{\frac{1}{s^2+4}\right\}\\ x\left(t\right)=\frac{1}{2}{L}^{-1}\left\{\frac{2}{s^2+2^2}\right\}\\ x\left(t\right)=\frac{1}{2}\left(\sin 2t\right)\\ x\left(t\right)=\frac{1}{2}\sin 2t\end{array}$

So, graph the solution $x\left(t\right)=\frac{1}{2}\sin 2t$ as shown below in Figure 1.

  

The solution of the differential equation is continuous for all real numbers.

Conclusion:

Thus, the solution of the given initial value problem is $x\left(t\right)=\frac{1}{2}\sin 2t$ and the graph of the solution is shown in Figure 1.