y''+y=e^{-2t}sin t , y(0)=y′(0)=0 solution of given initial value problem with Laplace transform

y''+y=e^{-2t}sin t , y(0)=y′(0)=0 solution of given initial value problem with Laplace transform

Question
Laplace transform
asked 2020-11-02
\(y''+y=e^{-2t}\sin t , y(0)=y′(0)=0\)
solution of given initial value problem with Laplace transform

Answers (1)

2020-11-03
Given:
\(y''+y=e^{-2t}\sin t , y(0)=y′(0)=0\)
To find: solution of given initial value problem with Laplace transform
Step 2 Solution
\(y''+y=e^{-2t}\sin t , y(0)=y′(0)=0\)
Taking Laplace transform of the differential equation , we get:
\([s^2Y(s)-sy(0)-y′(0)]+Y(s)= L[e^{-2t}\sin t]\) where \(Y(s) = L(y(t))\)
using initial values we get
\([s^2Y(s)-0]+Y(s) =L[e^{-2t}\sin t]\)
\(\text{Applying } L[e^{at}f(t)] = \bar f(s-a)\)
\(L(\sin t) = \frac{1}{s^2+1}\)
\(L(e^{-2t}\sin t) = \frac{1}{(s+2)^2 +1}\)
Step 3
\(\Rightarrow Y(s)(s^2+1) = \frac{1}{(s+2)^2+1}\)
\(\Rightarrow Y(s) = \frac{1}{(s^2+1)((s+2)^2 +1)}\)
after partial fraction we get
\(Y(s) = \frac{1}{8}\left[-\frac{s}{s^2+1}+\frac{1}{s^2+1} +\frac{s}{(s+2)^2 +1} +\frac{3}{(s+2)^2 +1}\right]\)
using inverse Laplace transform to get y(t) :
\(y(t) = L^{-1}[Y(s)]\)
\(=\frac{1}{8}\left[-\cos t + \sin t +e^{-2t}\cos t +3e^{-2t}\sin t\right]\)
required solution of differential equation .
0

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