Question

consider the piecewise differential equation
$x + 2x + 5x = \

$$\begin{cases} 17\\cos(2t) & \\text{if } 0 \\leqslant t \\leqslant 2\\pi \\\\ 0 & \\text{if } 2\\pi < t \\leqslant 6\\pi \\end{cases}$$

$
with initial conditions $x(0) = 6$ and $x(0) = 0$. both $x(t)$ and $x(t)$ are continuous on the interval $0, 6\pi$.

1. (1 point) rewrite the two pieces as a single equation using the heaviside step function $u(t)$. type your response by hand.

Explanation:

Step1: Recall the definition of Heaviside step function

The Heaviside step function \(u(t - a)\) is defined as \(u(t - a)=

$$\begin{cases}0, & t < a\\1, & t\geq a\end{cases}$$

\)

Step2: Express the right - hand side of the differential equation

For \(0\leq t\leq2\pi\), the function is \(17\cos(2t)\). For \(t > 2\pi\), the function is \(0\). We can write the right - hand side \(f(t)\) of the differential equation \(x'' + 2x'+5x=f(t)\) as:

$$
LATEXBLOCK1
$$

Answer:

\(x''+2x'+5x = 17\cos(2t)-17\cos(2t)u(t - 2\pi)\)