Inverse Laplace of $frac{s}{s^2+4s+5}$












0












$begingroup$


Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$



I obtained...



$$frac{s}{s^2+4s+4-4+5}$$



$$frac{s}{s^2+4s+1}$$



$$frac{s}{(s+2)^2+1}$$



I am not sure what my next step is?






ordinary-differential-equations







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$



    I obtained...



    $$frac{s}{s^2+4s+4-4+5}$$



    $$frac{s}{s^2+4s+1}$$



    $$frac{s}{(s+2)^2+1}$$



    I am not sure what my next step is?






    ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$



      I obtained...



      $$frac{s}{s^2+4s+4-4+5}$$



      $$frac{s}{s^2+4s+1}$$



      $$frac{s}{(s+2)^2+1}$$



      I am not sure what my next step is?






      ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$



      I obtained...



      $$frac{s}{s^2+4s+4-4+5}$$



      $$frac{s}{s^2+4s+1}$$



      $$frac{s}{(s+2)^2+1}$$



      I am not sure what my next step is?






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 5:50









      Eevee Trainer

      5,7961936




      5,7961936










      asked Dec 7 '18 at 5:48









      sumthatup11sumthatup11

      11




      11






















          1 Answer
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          5












          $begingroup$

          $$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$



          Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$



          $$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See mathworld.wolfram.com/LaplaceTransform.html
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 5:52










          • $begingroup$
            How did you get s = s+2 -2? Why did you add 2?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:00










          • $begingroup$
            @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 6:02












          • $begingroup$
            I do not understand your step in the numerator. I do not understand how s = s+2....?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:06











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          $$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$



          Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$



          $$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See mathworld.wolfram.com/LaplaceTransform.html
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 5:52










          • $begingroup$
            How did you get s = s+2 -2? Why did you add 2?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:00










          • $begingroup$
            @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 6:02












          • $begingroup$
            I do not understand your step in the numerator. I do not understand how s = s+2....?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:06
















          5












          $begingroup$

          $$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$



          Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$



          $$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See mathworld.wolfram.com/LaplaceTransform.html
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 5:52










          • $begingroup$
            How did you get s = s+2 -2? Why did you add 2?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:00










          • $begingroup$
            @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 6:02












          • $begingroup$
            I do not understand your step in the numerator. I do not understand how s = s+2....?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:06














          5












          5








          5





          $begingroup$

          $$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$



          Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$



          $$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$






          share|cite|improve this answer









          $endgroup$



          $$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$



          Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$



          $$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 5:51









          lab bhattacharjeelab bhattacharjee

          225k15157275




          225k15157275












          • $begingroup$
            See mathworld.wolfram.com/LaplaceTransform.html
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 5:52










          • $begingroup$
            How did you get s = s+2 -2? Why did you add 2?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:00










          • $begingroup$
            @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 6:02












          • $begingroup$
            I do not understand your step in the numerator. I do not understand how s = s+2....?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:06


















          • $begingroup$
            See mathworld.wolfram.com/LaplaceTransform.html
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 5:52










          • $begingroup$
            How did you get s = s+2 -2? Why did you add 2?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:00










          • $begingroup$
            @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
            $endgroup$
            – lab bhattacharjee
            Dec 7 '18 at 6:02












          • $begingroup$
            I do not understand your step in the numerator. I do not understand how s = s+2....?
            $endgroup$
            – sumthatup11
            Dec 7 '18 at 6:06
















          $begingroup$
          See mathworld.wolfram.com/LaplaceTransform.html
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 5:52




          $begingroup$
          See mathworld.wolfram.com/LaplaceTransform.html
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 5:52












          $begingroup$
          How did you get s = s+2 -2? Why did you add 2?
          $endgroup$
          – sumthatup11
          Dec 7 '18 at 6:00




          $begingroup$
          How did you get s = s+2 -2? Why did you add 2?
          $endgroup$
          – sumthatup11
          Dec 7 '18 at 6:00












          $begingroup$
          @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 6:02






          $begingroup$
          @sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
          $endgroup$
          – lab bhattacharjee
          Dec 7 '18 at 6:02














          $begingroup$
          I do not understand your step in the numerator. I do not understand how s = s+2....?
          $endgroup$
          – sumthatup11
          Dec 7 '18 at 6:06




          $begingroup$
          I do not understand your step in the numerator. I do not understand how s = s+2....?
          $endgroup$
          – sumthatup11
          Dec 7 '18 at 6:06


















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