Do I have something wrong when solving $y'+2y=6$?












5












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Solve $$y'+2y=6.$$




When I do $$y'=2(3-y)impliesintfrac{mathrm dy}{3-y}=2intmathrm dximplies-ln{|3-y|}=2x+cimplies3-y=ke^{-2x}therefore y=boxed{3-ke^{-2x}},quad c,kinBbb R.$$ It satisfies the ODE because $$2ke^{-2x}+6-2ke^{-2x}=6=6.$$ However, when I try another solution, namely first solve the homogeneous equation: $$y'+2y=0impliesintfrac{mathrm dy}y=-2intmathrm dximpliesln{|y|}=-2x+cimplies y=ke^{-2x},quad c,kinBbb R$$ then $y_P=k(x)e^{-2x}$, so then $$y'_P=k'(x)e^{-2x}-2k(x)e^{-2x}implies k'(x)e^{-2x}-2k(x)e^{-2x}+2k(x)e^{-2x}=6implies k'(x)=6e^{2x}implies k(x)=3e^{2x}implies y_P=3e^{2x}e^{-2x}=3therefore y=y_H+y_P=boxed{3+ke^{-2x}},$$ where this solution also satisfies $y'+2y=6$, because $$-2ke^{-2x}+6+2ke^{-2x}=6=6.$$ My question is, how can we express both solutions with the same expression of $y$? I would like both solutions to be identical, but how?



Thanks!






ordinary-differential-equations







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    5












    $begingroup$


    Solve $$y'+2y=6.$$




    When I do $$y'=2(3-y)impliesintfrac{mathrm dy}{3-y}=2intmathrm dximplies-ln{|3-y|}=2x+cimplies3-y=ke^{-2x}therefore y=boxed{3-ke^{-2x}},quad c,kinBbb R.$$ It satisfies the ODE because $$2ke^{-2x}+6-2ke^{-2x}=6=6.$$ However, when I try another solution, namely first solve the homogeneous equation: $$y'+2y=0impliesintfrac{mathrm dy}y=-2intmathrm dximpliesln{|y|}=-2x+cimplies y=ke^{-2x},quad c,kinBbb R$$ then $y_P=k(x)e^{-2x}$, so then $$y'_P=k'(x)e^{-2x}-2k(x)e^{-2x}implies k'(x)e^{-2x}-2k(x)e^{-2x}+2k(x)e^{-2x}=6implies k'(x)=6e^{2x}implies k(x)=3e^{2x}implies y_P=3e^{2x}e^{-2x}=3therefore y=y_H+y_P=boxed{3+ke^{-2x}},$$ where this solution also satisfies $y'+2y=6$, because $$-2ke^{-2x}+6+2ke^{-2x}=6=6.$$ My question is, how can we express both solutions with the same expression of $y$? I would like both solutions to be identical, but how?



    Thanks!






    ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      1



      $begingroup$


      Solve $$y'+2y=6.$$




      When I do $$y'=2(3-y)impliesintfrac{mathrm dy}{3-y}=2intmathrm dximplies-ln{|3-y|}=2x+cimplies3-y=ke^{-2x}therefore y=boxed{3-ke^{-2x}},quad c,kinBbb R.$$ It satisfies the ODE because $$2ke^{-2x}+6-2ke^{-2x}=6=6.$$ However, when I try another solution, namely first solve the homogeneous equation: $$y'+2y=0impliesintfrac{mathrm dy}y=-2intmathrm dximpliesln{|y|}=-2x+cimplies y=ke^{-2x},quad c,kinBbb R$$ then $y_P=k(x)e^{-2x}$, so then $$y'_P=k'(x)e^{-2x}-2k(x)e^{-2x}implies k'(x)e^{-2x}-2k(x)e^{-2x}+2k(x)e^{-2x}=6implies k'(x)=6e^{2x}implies k(x)=3e^{2x}implies y_P=3e^{2x}e^{-2x}=3therefore y=y_H+y_P=boxed{3+ke^{-2x}},$$ where this solution also satisfies $y'+2y=6$, because $$-2ke^{-2x}+6+2ke^{-2x}=6=6.$$ My question is, how can we express both solutions with the same expression of $y$? I would like both solutions to be identical, but how?



      Thanks!






      ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      Solve $$y'+2y=6.$$




      When I do $$y'=2(3-y)impliesintfrac{mathrm dy}{3-y}=2intmathrm dximplies-ln{|3-y|}=2x+cimplies3-y=ke^{-2x}therefore y=boxed{3-ke^{-2x}},quad c,kinBbb R.$$ It satisfies the ODE because $$2ke^{-2x}+6-2ke^{-2x}=6=6.$$ However, when I try another solution, namely first solve the homogeneous equation: $$y'+2y=0impliesintfrac{mathrm dy}y=-2intmathrm dximpliesln{|y|}=-2x+cimplies y=ke^{-2x},quad c,kinBbb R$$ then $y_P=k(x)e^{-2x}$, so then $$y'_P=k'(x)e^{-2x}-2k(x)e^{-2x}implies k'(x)e^{-2x}-2k(x)e^{-2x}+2k(x)e^{-2x}=6implies k'(x)=6e^{2x}implies k(x)=3e^{2x}implies y_P=3e^{2x}e^{-2x}=3therefore y=y_H+y_P=boxed{3+ke^{-2x}},$$ where this solution also satisfies $y'+2y=6$, because $$-2ke^{-2x}+6+2ke^{-2x}=6=6.$$ My question is, how can we express both solutions with the same expression of $y$? I would like both solutions to be identical, but how?



      Thanks!






      ordinary-differential-equations




      ordinary-differential-equations






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      asked Jan 15 at 0:26









      manoooohmanooooh

      6651518




      6651518






















          3 Answers
          3






          active

          oldest

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          4












          $begingroup$

          Both solutions are the same, $k$ is a real number, you can write $3+(-k)e^{-2x}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh I forgot that, thank you!!
            $endgroup$
            – manooooh
            Jan 15 at 0:32



















          2












          $begingroup$

          They are identical. You have no reason to expect $k $ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
            $endgroup$
            – manooooh
            Jan 15 at 0:34



















          1












          $begingroup$

          To see both are the same:
          $$-ln|3-y|=2x+A Rightarrow ln|3-y|=-2x-A Rightarrow |3-y|=e^{-2x-A} Rightarrow |3-y|=e^{-A}e^{-2x} Rightarrow 3-y=-Ce^{-2x} Rightarrow y=3+Ce^{-2x}.$$
          Alternatively (instead of variations):
          $$y=y_h+y_p=Ce^{-2x}+A;\
          [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 Rightarrow \
          2A=6 Rightarrow A=3 Rightarrow \
          y=Ce^{-2x}+3.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
            $endgroup$
            – manooooh
            Jan 15 at 1:15














          Your Answer








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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Both solutions are the same, $k$ is a real number, you can write $3+(-k)e^{-2x}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh I forgot that, thank you!!
            $endgroup$
            – manooooh
            Jan 15 at 0:32
















          4












          $begingroup$

          Both solutions are the same, $k$ is a real number, you can write $3+(-k)e^{-2x}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh I forgot that, thank you!!
            $endgroup$
            – manooooh
            Jan 15 at 0:32














          4












          4








          4





          $begingroup$

          Both solutions are the same, $k$ is a real number, you can write $3+(-k)e^{-2x}$.






          share|cite|improve this answer











          $endgroup$



          Both solutions are the same, $k$ is a real number, you can write $3+(-k)e^{-2x}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 4:41









          Moo

          5,68531020




          5,68531020










          answered Jan 15 at 0:31









          Tsemo AristideTsemo Aristide

          60.9k11446




          60.9k11446












          • $begingroup$
            Oh I forgot that, thank you!!
            $endgroup$
            – manooooh
            Jan 15 at 0:32


















          • $begingroup$
            Oh I forgot that, thank you!!
            $endgroup$
            – manooooh
            Jan 15 at 0:32
















          $begingroup$
          Oh I forgot that, thank you!!
          $endgroup$
          – manooooh
          Jan 15 at 0:32




          $begingroup$
          Oh I forgot that, thank you!!
          $endgroup$
          – manooooh
          Jan 15 at 0:32











          2












          $begingroup$

          They are identical. You have no reason to expect $k $ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
            $endgroup$
            – manooooh
            Jan 15 at 0:34
















          2












          $begingroup$

          They are identical. You have no reason to expect $k $ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
            $endgroup$
            – manooooh
            Jan 15 at 0:34














          2












          2








          2





          $begingroup$

          They are identical. You have no reason to expect $k $ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.






          share|cite|improve this answer









          $endgroup$



          They are identical. You have no reason to expect $k $ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 0:32









          Martin ArgeramiMartin Argerami

          130k1184185




          130k1184185












          • $begingroup$
            There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
            $endgroup$
            – manooooh
            Jan 15 at 0:34


















          • $begingroup$
            There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
            $endgroup$
            – manooooh
            Jan 15 at 0:34
















          $begingroup$
          There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
          $endgroup$
          – manooooh
          Jan 15 at 0:34




          $begingroup$
          There is indeed an initial condition, but I wanted to know why I end up with two "different" solutions. Thanks!
          $endgroup$
          – manooooh
          Jan 15 at 0:34











          1












          $begingroup$

          To see both are the same:
          $$-ln|3-y|=2x+A Rightarrow ln|3-y|=-2x-A Rightarrow |3-y|=e^{-2x-A} Rightarrow |3-y|=e^{-A}e^{-2x} Rightarrow 3-y=-Ce^{-2x} Rightarrow y=3+Ce^{-2x}.$$
          Alternatively (instead of variations):
          $$y=y_h+y_p=Ce^{-2x}+A;\
          [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 Rightarrow \
          2A=6 Rightarrow A=3 Rightarrow \
          y=Ce^{-2x}+3.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
            $endgroup$
            – manooooh
            Jan 15 at 1:15


















          1












          $begingroup$

          To see both are the same:
          $$-ln|3-y|=2x+A Rightarrow ln|3-y|=-2x-A Rightarrow |3-y|=e^{-2x-A} Rightarrow |3-y|=e^{-A}e^{-2x} Rightarrow 3-y=-Ce^{-2x} Rightarrow y=3+Ce^{-2x}.$$
          Alternatively (instead of variations):
          $$y=y_h+y_p=Ce^{-2x}+A;\
          [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 Rightarrow \
          2A=6 Rightarrow A=3 Rightarrow \
          y=Ce^{-2x}+3.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
            $endgroup$
            – manooooh
            Jan 15 at 1:15
















          1












          1








          1





          $begingroup$

          To see both are the same:
          $$-ln|3-y|=2x+A Rightarrow ln|3-y|=-2x-A Rightarrow |3-y|=e^{-2x-A} Rightarrow |3-y|=e^{-A}e^{-2x} Rightarrow 3-y=-Ce^{-2x} Rightarrow y=3+Ce^{-2x}.$$
          Alternatively (instead of variations):
          $$y=y_h+y_p=Ce^{-2x}+A;\
          [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 Rightarrow \
          2A=6 Rightarrow A=3 Rightarrow \
          y=Ce^{-2x}+3.$$






          share|cite|improve this answer









          $endgroup$



          To see both are the same:
          $$-ln|3-y|=2x+A Rightarrow ln|3-y|=-2x-A Rightarrow |3-y|=e^{-2x-A} Rightarrow |3-y|=e^{-A}e^{-2x} Rightarrow 3-y=-Ce^{-2x} Rightarrow y=3+Ce^{-2x}.$$
          Alternatively (instead of variations):
          $$y=y_h+y_p=Ce^{-2x}+A;\
          [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 Rightarrow \
          2A=6 Rightarrow A=3 Rightarrow \
          y=Ce^{-2x}+3.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 1:08









          farruhotafarruhota

          22.5k2942




          22.5k2942












          • $begingroup$
            Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
            $endgroup$
            – manooooh
            Jan 15 at 1:15




















          • $begingroup$
            Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
            $endgroup$
            – manooooh
            Jan 15 at 1:15


















          $begingroup$
          Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
          $endgroup$
          – manooooh
          Jan 15 at 1:15






          $begingroup$
          Oh, you did $e^{-A}to-C$ when $|3-y|to(3-y)$ instead of $e^{-A}to C$, that is nice! Also I like your approach $ddotsmile$.
          $endgroup$
          – manooooh
          Jan 15 at 1:15




















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