Question

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a delta function. y′+y=7+δ(t−3),y(0)=0. y′+y=7+δ(t−3),y(0)=0. Find the Laplace transform of the solution. Y(s)=L{y(t)}=Y(s)=L{y(t)}= Obtain the solution y(t)y(t).

Answer 1

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying  laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get  $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$