Question

In Problems 63–70, use the Laplace transform to solve the given initial-value problem.

$y+4y=f\left(t\right),y\left(0\right)=0,y\left(0\right)=-1wherewheref\left(t\right)=\left\{\begin{array}{ll}1,& 0⩽t<1\\ 0,& t⩾1\end{array}\right.$

Short answer

The Laplace transform of the given function is

$y\left(t\right)=\frac{1}{4}-\frac{1}{4}\cos \left(2t\right)-\frac{1}{4}U(t-1)+\frac{1}{4}\cos \left(2\right(t-1\left)\right)U(t-1)-\frac{1}{2}\sin \left(2t\right)$

Step-by-step solution

Definition of Laplace Transform

• Laplace transform is the integral transform of the given derivative function with real variable t to convert into a complex function with variables.
• The Laplace transform of f(t) that is denoted by $L\left\{f\left(t\right)\right\}$or $F\left(s\right)$is defined by the Laplace transform formula,
• $F\left(s\right)={\int }_0^{+\infty }f\left(t\right)·e^{-s·t}·dt$
  

Find the value of Y 

Given that,

$L\left\{y\text{'}\text{'}+4\text{y}\right\}=L\left\{f\left(t\right)\right\}\dots \dots y\left(0\right)=0$

Where,

$$\begin{gathered} f\left(t\right)=\left\{\begin{array}{ll}1,& 0⩽t<1\\ 0,& t⩾1\end{array}\right. \\ f\left(t\right)=1-U(t-1) \\ {s}^{2}Y\left(s\right)-s\left(0\right)-(-1)+4Y=\frac{1}{s}-\frac{e^{-s}}{s} \\ Y\left(s\right)=\frac{1}{s\left(s^2+2\right)}-\frac{e^{-s}}{s\left(s^2+2\right)}-\frac{e^{-s}}{s^2+4} \end{gathered}$$

Decomposing

$$\begin{gathered} \frac{1}{s^2(s+2)}=\frac{a}{s}+\frac{bs+c}{s^2+4} \\ =\frac{1}{4s}-\frac{s}{4\left(s^2+4\right)} \\ Y=\frac{1}{4s}-\frac{s}{4\left(s^2+4\right)}-\frac{e^{-s}}{4s}+\frac{s{e}^{-s}}{4\left(s^2+4\right)}-\frac{1}{2}\frac{2}{s^2+4} \end{gathered}$$

Unit Step Function

The unit step function $u(t-a)$is defined to b

$$\begin{gathered} u\left(t-a\right)=\left\{\begin{array}{ll}0,& 0⩽t<1\\ 1,& t⩾a\end{array}\right. \\ y\left(t\right)=L^{-1}\left\{\frac{1}{4s}-\frac{s}{4\left(s^2+4\right)}-\frac{e^{-s}}{4s}+\frac{s{e}^{-s}}{4\left(s^2+4\right)}-\frac{1}{2}\frac{2}{s^2+4}\right\} \\ y\left(t\right)=\frac{1}{4}-\frac{1}{4}\cos \left(2t\right)-\frac{1}{4}U(t-1)+\frac{1}{4}\cos \left(2\right(t-1\left)\right)U(t-1)-\frac{1}{2}\sin \left(2t\right) \end{gathered}$$