Let y(t)=int_0^tf(t)dt If the Laplace transform of y(t) is given Y(s)=frac{19}{(s^2+25)} , find f(t) a) f(t)=19 sin(5t) b) none c) f(t)=6 sin(2t) d) f(t)=20 cos(6t) e) f(t)=19 cos(5t)

Armorikam 2021-01-15 Answered
Let y(t)=0tf(t)dt If the Laplace transform of y(t) is given Y(s)=19(s2+25) , find f(t)
a) f(t)=19sin(5t)
b) none
c) f(t)=6sin(2t)
d) f(t)=20cos(6t)
e) f(t)=19cos(5t)
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

funblogC
Answered 2021-01-16 Author has 91 answers
Step 1
The Laplace transform of y(t) is given Y(s)=19(s2+25)
The Laplace transform is written as,
L{f(t)}=F(s)
L[0tf(t)dt]=1sF(s)
It is given that , y(t)=0tf(t)dt
Take Laplace on both sides,
L{y(t)}=L{0tf(t)dt}
Step 2
Substitute Laplace of y(t) ,
1sL{f(t)}=19(s2+25)
Lf(t)=19s(s2+25)
Take inverse on both sides,
f(t)=L1[19(s2+25)]
=19L1[s(s2+25)]
The inverse Laplace transform of cos(at) is L1[s(s2+a2)]
Step 3
Therefore ,
f(t)=L1[19s(s2+25)]
=19L1[s(s2+25)]
=19cos5t
The value of f(t)=19cos(5t) and the correct option is E.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more