Method 1 (Using Shifting Property)
L−1{s2+2s−311−3s}=L−1{s2+2s+1−4−3(s+1)+14}=L−1{(s+1)2−4−3(s+1)+14}=e−tL−1{s2−4−3s+14}=e−t[L−1{s2−4−3s}+L−1{s2−414}]=e−t[−3cosh2t+7sinh2t]=e−t[−3(2et+e−t)+7(2et−e−t)]=2e−t[−3(e2t+e−2t)+7(e2t−e−2t)]=2e−t[4e2t−10e−2t]=2et−5e−3t
Method 2 (Partial fraction method)
We use partial fraction to split s2+2s−311−3s.
s2+2s−311−3s11−3s=s+3A+s−1B=A(s−1)+B(s+3)Taking s=1⟹8=4B⟹B=2 Taking s=−3⟹20=−4A⟹A=−5 ∴s2+2s−311−3s=s+3−5+s−12
L−1{s2+2s−311−3s}=L−1{s+3−5+s−12}=2et−5e−3t
From the above two methods, we see that the answers are equal.
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