$(D^5-D)y = 12e^x$ using exponential shift












0












$begingroup$


The equation is $$(D^5-D)y = 12e^x$$



Here, the general solution of homogenous (D^5-D)y=0 is $y_g = c_1 + c_2 cos x + c_3 sin x + c_4 e^x + c_5 e^{-x}$. Now I can apply two methods:




  1. Assume particular solution to be $Axe^x$ and solve. This is linearly independant from $y_g$. But finding $D^5$ can be hectic.


  2. Use exponential shift:



$$(D^5-D)y = 12e^x\
y = 12 e^xfrac{1}{(D+1)^5-(D+1)}(1) \
= 12 e^xfrac{1}{(D+1)^4-1}(1)\
= 12e^x (1+(D+1)^4 + ....)$$



This gives strange answer. What is the proper method for this?






calculus ordinary-differential-equations







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    The equation is $$(D^5-D)y = 12e^x$$



    Here, the general solution of homogenous (D^5-D)y=0 is $y_g = c_1 + c_2 cos x + c_3 sin x + c_4 e^x + c_5 e^{-x}$. Now I can apply two methods:




    1. Assume particular solution to be $Axe^x$ and solve. This is linearly independant from $y_g$. But finding $D^5$ can be hectic.


    2. Use exponential shift:



    $$(D^5-D)y = 12e^x\
    y = 12 e^xfrac{1}{(D+1)^5-(D+1)}(1) \
    = 12 e^xfrac{1}{(D+1)^4-1}(1)\
    = 12e^x (1+(D+1)^4 + ....)$$



    This gives strange answer. What is the proper method for this?






    calculus ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The equation is $$(D^5-D)y = 12e^x$$



      Here, the general solution of homogenous (D^5-D)y=0 is $y_g = c_1 + c_2 cos x + c_3 sin x + c_4 e^x + c_5 e^{-x}$. Now I can apply two methods:




      1. Assume particular solution to be $Axe^x$ and solve. This is linearly independant from $y_g$. But finding $D^5$ can be hectic.


      2. Use exponential shift:



      $$(D^5-D)y = 12e^x\
      y = 12 e^xfrac{1}{(D+1)^5-(D+1)}(1) \
      = 12 e^xfrac{1}{(D+1)^4-1}(1)\
      = 12e^x (1+(D+1)^4 + ....)$$



      This gives strange answer. What is the proper method for this?






      calculus ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      The equation is $$(D^5-D)y = 12e^x$$



      Here, the general solution of homogenous (D^5-D)y=0 is $y_g = c_1 + c_2 cos x + c_3 sin x + c_4 e^x + c_5 e^{-x}$. Now I can apply two methods:




      1. Assume particular solution to be $Axe^x$ and solve. This is linearly independant from $y_g$. But finding $D^5$ can be hectic.


      2. Use exponential shift:



      $$(D^5-D)y = 12e^x\
      y = 12 e^xfrac{1}{(D+1)^5-(D+1)}(1) \
      = 12 e^xfrac{1}{(D+1)^4-1}(1)\
      = 12e^x (1+(D+1)^4 + ....)$$



      This gives strange answer. What is the proper method for this?






      calculus ordinary-differential-equations




      calculus ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 19 at 6:49









      jeeajeea

      58515




      58515






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          1. More efficient way is as follows. Let us write $T=D^5-D$. For any $rinBbb C$, we have
          $$
          T[e^{rx}]=frac{d^5}{dx^5}e^{rx}-frac{d}{dx}e^{rx}=(r^5-r)e^{rx}.
          $$           Now, differentiate with respect to $r$. Then we get
          $$
          frac{partial }{partial r}T[e^{rx}]=Tleft[frac{partial }{partial r}e^{rx}right]=T[xe^{rx}]=frac{partial }{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
          $$           We get by letting $r=1$
          $$
          T[xe^x]=4e^x.
          $$           Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.



          2. In fact, correct formula should be
          $$
          y=12(D^5-D)^{-1}[e^{x}],
          $$           not
          $$
          y=12e^x (D^5-D)^{-1}[1].
          $$
          Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have
          $$
          (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x.
          $$           Finally, since $(D-1)[xe^x]=e^x$, we have
          $$
          (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x.
          $$           This also gives a particular solution $3xe^x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            sorry i dont understand your method 2
            $endgroup$
            – jeea
            Jan 19 at 16:38










          • $begingroup$
            $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
            $endgroup$
            – Song
            Jan 19 at 16:42












          • $begingroup$
            Yes the first line only, from 12ex to 3ex
            $endgroup$
            – jeea
            Jan 19 at 17:27






          • 1




            $begingroup$
            @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
            $endgroup$
            – Song
            Jan 19 at 17:41












          • $begingroup$
            Oh so we have to observe and predict very carefullly
            $endgroup$
            – jeea
            Jan 20 at 4:20











          Your Answer





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          active

          oldest

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          1












          $begingroup$

          1. More efficient way is as follows. Let us write $T=D^5-D$. For any $rinBbb C$, we have
          $$
          T[e^{rx}]=frac{d^5}{dx^5}e^{rx}-frac{d}{dx}e^{rx}=(r^5-r)e^{rx}.
          $$           Now, differentiate with respect to $r$. Then we get
          $$
          frac{partial }{partial r}T[e^{rx}]=Tleft[frac{partial }{partial r}e^{rx}right]=T[xe^{rx}]=frac{partial }{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
          $$           We get by letting $r=1$
          $$
          T[xe^x]=4e^x.
          $$           Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.



          2. In fact, correct formula should be
          $$
          y=12(D^5-D)^{-1}[e^{x}],
          $$           not
          $$
          y=12e^x (D^5-D)^{-1}[1].
          $$
          Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have
          $$
          (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x.
          $$           Finally, since $(D-1)[xe^x]=e^x$, we have
          $$
          (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x.
          $$           This also gives a particular solution $3xe^x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            sorry i dont understand your method 2
            $endgroup$
            – jeea
            Jan 19 at 16:38










          • $begingroup$
            $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
            $endgroup$
            – Song
            Jan 19 at 16:42












          • $begingroup$
            Yes the first line only, from 12ex to 3ex
            $endgroup$
            – jeea
            Jan 19 at 17:27






          • 1




            $begingroup$
            @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
            $endgroup$
            – Song
            Jan 19 at 17:41












          • $begingroup$
            Oh so we have to observe and predict very carefullly
            $endgroup$
            – jeea
            Jan 20 at 4:20
















          1












          $begingroup$

          1. More efficient way is as follows. Let us write $T=D^5-D$. For any $rinBbb C$, we have
          $$
          T[e^{rx}]=frac{d^5}{dx^5}e^{rx}-frac{d}{dx}e^{rx}=(r^5-r)e^{rx}.
          $$           Now, differentiate with respect to $r$. Then we get
          $$
          frac{partial }{partial r}T[e^{rx}]=Tleft[frac{partial }{partial r}e^{rx}right]=T[xe^{rx}]=frac{partial }{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
          $$           We get by letting $r=1$
          $$
          T[xe^x]=4e^x.
          $$           Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.



          2. In fact, correct formula should be
          $$
          y=12(D^5-D)^{-1}[e^{x}],
          $$           not
          $$
          y=12e^x (D^5-D)^{-1}[1].
          $$
          Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have
          $$
          (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x.
          $$           Finally, since $(D-1)[xe^x]=e^x$, we have
          $$
          (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x.
          $$           This also gives a particular solution $3xe^x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            sorry i dont understand your method 2
            $endgroup$
            – jeea
            Jan 19 at 16:38










          • $begingroup$
            $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
            $endgroup$
            – Song
            Jan 19 at 16:42












          • $begingroup$
            Yes the first line only, from 12ex to 3ex
            $endgroup$
            – jeea
            Jan 19 at 17:27






          • 1




            $begingroup$
            @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
            $endgroup$
            – Song
            Jan 19 at 17:41












          • $begingroup$
            Oh so we have to observe and predict very carefullly
            $endgroup$
            – jeea
            Jan 20 at 4:20














          1












          1








          1





          $begingroup$

          1. More efficient way is as follows. Let us write $T=D^5-D$. For any $rinBbb C$, we have
          $$
          T[e^{rx}]=frac{d^5}{dx^5}e^{rx}-frac{d}{dx}e^{rx}=(r^5-r)e^{rx}.
          $$           Now, differentiate with respect to $r$. Then we get
          $$
          frac{partial }{partial r}T[e^{rx}]=Tleft[frac{partial }{partial r}e^{rx}right]=T[xe^{rx}]=frac{partial }{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
          $$           We get by letting $r=1$
          $$
          T[xe^x]=4e^x.
          $$           Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.



          2. In fact, correct formula should be
          $$
          y=12(D^5-D)^{-1}[e^{x}],
          $$           not
          $$
          y=12e^x (D^5-D)^{-1}[1].
          $$
          Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have
          $$
          (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x.
          $$           Finally, since $(D-1)[xe^x]=e^x$, we have
          $$
          (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x.
          $$           This also gives a particular solution $3xe^x$.






          share|cite|improve this answer











          $endgroup$



          1. More efficient way is as follows. Let us write $T=D^5-D$. For any $rinBbb C$, we have
          $$
          T[e^{rx}]=frac{d^5}{dx^5}e^{rx}-frac{d}{dx}e^{rx}=(r^5-r)e^{rx}.
          $$           Now, differentiate with respect to $r$. Then we get
          $$
          frac{partial }{partial r}T[e^{rx}]=Tleft[frac{partial }{partial r}e^{rx}right]=T[xe^{rx}]=frac{partial }{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
          $$           We get by letting $r=1$
          $$
          T[xe^x]=4e^x.
          $$           Hence, $T[3xe^x]=12e^x$ and $3xe^x$ is a particular solution of the equation.



          2. In fact, correct formula should be
          $$
          y=12(D^5-D)^{-1}[e^{x}],
          $$           not
          $$
          y=12e^x (D^5-D)^{-1}[1].
          $$
          Note that $D^5-D=(D-1)(D^4+D^3+D^2+D)$. Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have
          $$
          (D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x.
          $$           Finally, since $(D-1)[xe^x]=e^x$, we have
          $$
          (D-1)^{-1}[3e^x]=(D-1)^{-1}(D^4+D^3+D^2+D)^{-1}[12e^x]=3xe^x.
          $$           This also gives a particular solution $3xe^x$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 19 at 10:00

























          answered Jan 19 at 7:56









          SongSong

          13.2k632




          13.2k632












          • $begingroup$
            sorry i dont understand your method 2
            $endgroup$
            – jeea
            Jan 19 at 16:38










          • $begingroup$
            $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
            $endgroup$
            – Song
            Jan 19 at 16:42












          • $begingroup$
            Yes the first line only, from 12ex to 3ex
            $endgroup$
            – jeea
            Jan 19 at 17:27






          • 1




            $begingroup$
            @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
            $endgroup$
            – Song
            Jan 19 at 17:41












          • $begingroup$
            Oh so we have to observe and predict very carefullly
            $endgroup$
            – jeea
            Jan 20 at 4:20


















          • $begingroup$
            sorry i dont understand your method 2
            $endgroup$
            – jeea
            Jan 19 at 16:38










          • $begingroup$
            $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
            $endgroup$
            – Song
            Jan 19 at 16:42












          • $begingroup$
            Yes the first line only, from 12ex to 3ex
            $endgroup$
            – jeea
            Jan 19 at 17:27






          • 1




            $begingroup$
            @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
            $endgroup$
            – Song
            Jan 19 at 17:41












          • $begingroup$
            Oh so we have to observe and predict very carefullly
            $endgroup$
            – jeea
            Jan 20 at 4:20
















          $begingroup$
          sorry i dont understand your method 2
          $endgroup$
          – jeea
          Jan 19 at 16:38




          $begingroup$
          sorry i dont understand your method 2
          $endgroup$
          – jeea
          Jan 19 at 16:38












          $begingroup$
          $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
          $endgroup$
          – Song
          Jan 19 at 16:42






          $begingroup$
          $$12e^x stackrel{(D^4+D^3+D^2+D)^{-1}}Longrightarrow 3e^xstackrel{(D-1)^{-1}}Longrightarrow 3xe^x.$$ $$12e^x stackrel{D^4+D^3+D^2+D}Longleftarrow 3e^xstackrel{D-1}Longleftarrow 3xe^x.$$Can you please be more specific about which part of the argument you don't understand?
          $endgroup$
          – Song
          Jan 19 at 16:42














          $begingroup$
          Yes the first line only, from 12ex to 3ex
          $endgroup$
          – jeea
          Jan 19 at 17:27




          $begingroup$
          Yes the first line only, from 12ex to 3ex
          $endgroup$
          – jeea
          Jan 19 at 17:27




          1




          1




          $begingroup$
          @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
          $endgroup$
          – Song
          Jan 19 at 17:41






          $begingroup$
          @jeea Since $(D^4+D^3+D^2+D)[e^x]=4e^x$, we have $(D^4+D^3+D^2+D)^{-1}[4e^x]=e^x$. By multiplying $3$, we have $(D^4+D^3+D^2+D)^{-1}[12e^x]=3e^x$.
          $endgroup$
          – Song
          Jan 19 at 17:41














          $begingroup$
          Oh so we have to observe and predict very carefullly
          $endgroup$
          – jeea
          Jan 20 at 4:20




          $begingroup$
          Oh so we have to observe and predict very carefullly
          $endgroup$
          – jeea
          Jan 20 at 4:20


















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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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          Antonio Litta Visconti Arese

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          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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