Question

Use the Laplace transform to solve the given initial-value problem.

$y^{\text{'}}+y=f\left(t\right),y\left(0\right)=0,wheref\left(t\right)=\left\{\begin{array}{ll}1,& 0⩽t<1\\ -1,& t⩾1\end{array}\right.$

Short answer

The initial value of the Laplace transform function is$y\left(t\right)=1-e^{-t}-2U(t-1)+2e^{-(t-1)}U(t-1)$

Step-by-step solution

Definition of "Laplace Transform"

• 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 supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by Lf(t) or F(s):

$F\left(s\right)={\int }_0^{+\infty }f\left(t\right)·e^{-s·t}·dt$

Find the Initial Value

Given that,

$$\begin{gathered} L\left\{y^{\text{'}}+y\right\}=L\left\{f\right(t\left)\right\}\dots \dots y\left(0\right)=0 \\ f\left(t\right)=1-2U(t-1) \end{gathered}$$

Where,

role="math" localid="1663924496016" $$\begin{gathered} sY-0+Y=\frac{1}{s}-\frac{2e^{-s}}{s} \\ Y=\frac{1}{s(s+1)}-\frac{2e^{-s}}{s(s+1)} \end{gathered}$$

1. Decomposing the equation

$\frac{1}{s(s+1)}=\frac{a}{s}+\frac{b}{s+1}=\frac{1}{s}-\frac{1}{s+1}$

2. Decomposing

$$\begin{gathered} \frac{2}{s(s+1)}=\frac{a}{s}+\frac{b}{s+1}=\frac{2}{s}-\frac{2}{s+1} \\ Y=\frac{1}{s}-\frac{1}{s+1}-\frac{2e^{-s}}{s}+\frac{2e^{-s}}{s+1} \end{gathered}$$

$$\begin{gathered} y\left(t\right)=L^1\left\{\frac{1}{s}-\frac{1}{s+1}-\frac{2e^{-s}}{s}+\frac{2e^{-s}}{s+1}\right\} \\ y\left(t\right)=1-e^{-t}-2U(t-1)+2e^{-(t-1)}U(t-1) \end{gathered}$$

Hence, the initial value of the Laplace transform function is

$y\left(t\right)=1-e^{-t}-2U(t-1)+2e^{-(t-1)}U(t-1)$