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

BenoguigoliB 2020-11-02 Answered
y+y=e2tsint,y(0)=y(0)=0
solution of given initial value problem with Laplace transform
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Expert Answer

Nathanael Webber
Answered 2020-11-03 Author has 117 answers
Given:
y+y=e2tsint,y(0)=y(0)=0
To find: solution of given initial value problem with Laplace transform
Step 2 Solution
y+y=e2tsint,y(0)=y(0)=0
Taking Laplace transform of the differential equation , we get:
[s2Y(s)sy(0)y(0)]+Y(s)=L[e2tsint] where Y(s)=L(y(t))
using initial values we get
[s2Y(s)0]+Y(s)=L[e2tsint]
Applying L[eatf(t)]=f¯(sa)
L(sint)=1s2+1
L(e2tsint)=1(s+2)2+1
Step 3
Y(s)(s2+1)=1(s+2)2+1
Y(s)=1(s2+1)((s+2)2+1)
after partial fraction we get
Y(s)=18[ss2+1+1s2+1+s(s+2)2+1+3(s+2)2+1]
using inverse Laplace transform to get y(t) :
y(t)=L1[Y(s)]
=18[cost+sint+e2tcost+3e2tsint]
required solution of differential equation .
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