Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2

Name: Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2

Question

Find the inverse Laplace transform of F(s)=1(s+a)2(s+b)  for  t≥0  a) sin⁡(at)+e−bt  b) te−bt(a−b)2+sin⁡(at)  c) sin⁡(at)(b−a)2+e−bt(a−b)2  d) e−bt(a−b)2+te−atb−a−e−at(b−a)2  e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2

Pohanginah · Accepted Answer

Step 1   Here,   F(s)=1(s+a)2(s+b)  step 2   Rewrite the function.   F(s)=1a−b×(s+a)−(s+b)(s+a)2(s+b)  =1a−b×[(s+a)(s+a)2(s+b)−(s+b)(s+a)2(s+b)]  =1a−b×[1(s+a)(s+b)−1(s+a)2]  =1a−b×[1a−b×(s+a)−(s+b)(s+a)(s+b)−1!(s+a)1+1]  =1(a−b)2[1s+b−1s+a]−1a−b×1!(s+a)1+1  step 3   Apply inverse laplace transform formula.   f(t)=1(a−b)2[e−bt−e−at]−1a−b×t1e−at  =e−bt(a−b)2+te−atb−a−e−at(b−a)2  Step 4   Answer: Option D is correct.

Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at)...

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