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In Problems, determineL-1Fs2Fs-4Fs=5s+1

Short Answer

Expert verified

L-15s2-4s+1=512e2t+54e-2t-53e-t

Step by step solution

01

Find the factor of the denominator

From the given equation we get:

Fs=5s2-4s+1=5s-2s+2s+1

Using partial fractions, we get:

5s-2s+1s-1=As-2+Bs+2+Cs+1

from where it follows:

5=As+2s+1+Bs-2s+1+Cs2-4

Using s=2,-2,-1,respectively we get:

s=2:5=12A⇒A=512s=-2:5B=4⇒B=54s=-1:5=-3C⇒C=-53Therefore,5s2-4s+1=512s-2+54s+2-53s+1

02

Find the inverse

Find the inverse Laplace transform we get:

L-15s2-4s+1=L-1512s-2+L-154s+2+L-153s+1=512L-11s-2+54L-11s+2-53L-11s+1=512e2t+54e-2t-53e-tTherefore,L-15s2-4s+1=512e2t+54e-2t-53e-t

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