Mathematics, 21.06.2020 02:57 jenny8460
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).
Answers: 1
![L[{y+y'} ]= L[{7 + \delta (t-3)}]](/tpl/images/0691/2533/10733.png)
![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}]](/tpl/images/0691/2533/ef1e6.png)
![L^{-1}[\dfrac{e^{-3s}}{s+1}]](/tpl/images/0691/2533/d57de.png)
![L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;](/tpl/images/0691/2533/583ff.png)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t3} \atop {0 \ \ \ \ \ \ \ \ \ t](/tpl/images/0691/2533/ec57d.png)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t3} \atop {0 \ \ \ \ \ \ \ \ \ t](/tpl/images/0691/2533/1d539.png)
![y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](/tpl/images/0691/2533/610e4.png)
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Consider the following initial value problem, in which an input of large amplitude and short duratio...
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