Find inverse laplace transform of 2 ( ( s − 1 ) 2 + 1 ) 2 I tried to decompose the fraction using ∵ ( s − 1 ) 2 + 1 = s 2 − 2 s + 2 → 2 ( ( s − 1 ) 2 + 1 ) 2 = A s + B s 2 − 2 s + 2 + C s + D ( s 2 − 2 s + 2 ) 2 yet I get D=2, which leads me back to the same exact equation, any help?

Name: Find inverse laplace transform of 2 ( ( s − 1 ) 2 + 1 ) 2 I tried to decompose the fraction using ∵ ( s − 1 ) 2 + 1 = s 2 − 2 s + 2 → 2 ( ( s − 1 ) 2 + 1 ) 2 = A s + B s 2 − 2 s + 2 + C s + D ( s 2 − 2 s + 2 ) 2 yet I get D=2, which leads me back to the same exact equation, any help?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find inverse laplace transform of 2 ( ( s − 1 ) 2 + 1 ) 2 I tried to decompose the fraction using ∵ ( s − 1 ) 2 + 1 = s 2 − 2 s + 2 → 2 ( ( s − 1 ) 2 + 1 ) 2 = A s + B s 2 − 2 s + 2 + C s + D ( s 2 − 2 s + 2 ) 2 yet I get D=2, which leads me back to the same exact equation, any help?

Question

Find inverse laplace transform of      2          (      (      s      −      1              )        2            +      1              )        2            I tried to decompose the fraction using  ∵  (  s  −  1      )    2    +  1  =      s    2    −  2  s  +  2  →      2          (      (      s      −      1              )        2            +      1              )        2              =            A      s      +      B                      s        2            −      2      s      +      2        +            C      s      +      D              (              s        2            −      2      s      +      2              )        2            yet I get D=2, which leads me back to the same exact equation, any help?

yen1291kp6 · Accepted Answer

While there is a defined inverse Laplace transform, the integral is often difficult. Usually, it is easier to take the transforms we know and noodle with them until we find a fit.  L  {      e    t    cos  ⁡  t  }  =            s      −      1              (      s      −      1              )        2            +      1          L  {      e    t    sin  ⁡  t  }  =      1          (      s      −      1              )        2            +      1        =                    s        2            −      2      s      +      2              (      (      s      −      1              )        2            +      1              )        2                L  {  t      e    t    cos  ⁡  t  }  =  −      d          d      s                  (      s      −      1      )              (      s      −      1              )        2            +      1        =                    s        2            −      2      s              (      (      s      −      1              )        2            +      1              )        2                L  {  t      e    t    sin  ⁡  t  }  =  −      d          d      s            1          (      s      −      1              )        2            +      1        =            2      s      −      2              (      (      s      −      1              )        2            +      1              )        2            Some combination of   (      s    2    −  2  s  +  2  )  ,  (      s    2    −  2  s  )  ,  (  2  s  −  2  )  =  2      e    t    sin  ⁡  t  −  t      e    t    cos  ⁡  t

paratusojitos0yx · Accepted Answer

HINT:Use the following properties      L    {  t  f  (  t  )  }  (  s  )  =  −      d          d      s            L    {  f  (  t  )  }  (  s  )      L    {      e                  s        0            t        f  (  t  )  }  (  s  )  =      L    {  f  (  t  )  }  (  s  −      s    0    )            along with these Laplace Transform pairs      L    {  cos  ⁡  (  t  )  }  (  s  )  =      s                  s        2            +      1            L    {  sin  ⁡  (  t  )  }  (  s  )  =      1                  s        2            +      1

Find inverse laplace transform of (2)/(((s−1)^2+1)^2)

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