Find L^{-1}left{frac{1}{(s + 6)(s - 4)}right}

Daniaal Sanchez 2020-12-29 Answered
Find L1{1(s + 6)(s  4)}
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

Raheem Donnelly
Answered 2020-12-30 Author has 75 answers

Recall: Linearity Property: Suppose f1(p) and  f2(p) are Laplace transformation
of F1(t) and  F2(t) respectively. Then,
L{c1F1(t) ± c2F2(t)}=c1L{F1(t)} ± c2L{F2(t)}=c1f1(p) ± c2f2(p).
For Inverse Laplace transformation,
L1{c1f1(p) ± c2f2(p)}=cL1{f1(p)} ± c2L1{f2(p)}=c  1F1(t) ± c2F2(t).
Results:
L1{1s  a}=cat.
We have to find the value of L1{1(s + 6)(s  4)}.
1(s + 6)(s  4) Can be written as,
1(s + 6)(s  4)=A(s + 6) B(s  4)
=(A + B)s + (4A + 6B)(s + 6)(s  4)
Comparing the coefficients of s and the constant term, we get,
A + B=0 6B  4A=1
 A= B  10B=1[putting A= B]
A= 110  B=110
So, 1(s + 6)(s  4)=110 (1s  4))  110(1s + 6)
Taking Inverse Laplace transformation on both sides of the above equation, we get,
L1{1(s + 6)(s  4)}=L1{110(1s  4) 110(1s + 6)}
=110L1{1s  4}  110L1{1s + 6}
=110c4t  110c6t
=c4t  c6t10
Therefore, the required solution is,
L1

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