$(D^5-D)y = 12e^x$ 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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          The Binding of Isaac: Rebirth/Afterbirth

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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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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