1
Question
Consider the differential equationd2y(t)dt2+2dy(t)dt+y(t)=δ(t) with y(t)|t=0=−2 anddydt|t=0=0.. The numerical value of dydt|t=0 is
Consider the differential equationd2y(t)dt2+2dy(t)dt+y(t)=δ(t) with y(t)|t=0=−2 anddydt|t=0=0.. The numerical value of dydt|t=0 is
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Solution
The correct option is D 1
Given D.E. can be written as
y′′+2y′+y=δ(t)
Taking L.T both sides
L{y"}+2L{y'}+L{y} =L{δ(t)}
s2¯y−sy(0)−y′(0)]+2[s¯y−y(0)]+¯y=1
(s2+2s+1)¯y+2s+4=1
¯y=−3−2s(s+1)2=−3(s+1)2−2s(s+1)2
=−3(s+1)2−2(s+1)−1(s+1)2
=−3(s+1)2−2(s+1)+2(s+1)2
So y=L−1{¯y}=−3te−t−2e−t+2te−ty(t)=−te−t−2e−t
It is the solution of given Differential Equaltion
Now, dydt=te−t−e−t+2e−t
So, At t=0,dydt=0−1+2=1
Given D.E. can be written as
y′′+2y′+y=δ(t)
Taking L.T both sides
L{y"}+2L{y'}+L{y} =L{δ(t)}
s2¯y−sy(0)−y′(0)]+2[s¯y−y(0)]+¯y=1
(s2+2s+1)¯y+2s+4=1
¯y=−3−2s(s+1)2=−3(s+1)2−2s(s+1)2
=−3(s+1)2−2(s+1)−1(s+1)2
=−3(s+1)2−2(s+1)+2(s+1)2
So y=L−1{¯y}=−3te−t−2e−t+2te−ty(t)=−te−t−2e−t
It is the solution of given Differential Equaltion
Now, dydt=te−t−e−t+2e−t
So, At t=0,dydt=0−1+2=1
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