Finding the complementary function of a second order linear differential equation












1












$begingroup$


I have a question stating




Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$




So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?



EDIT



To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?






calculus ordinary-differential-equations complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
    $endgroup$
    – LutzL
    Feb 26 '18 at 17:34
















1












$begingroup$


I have a question stating




Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$




So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?



EDIT



To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?






calculus ordinary-differential-equations complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
    $endgroup$
    – LutzL
    Feb 26 '18 at 17:34














1












1








1


1



$begingroup$


I have a question stating




Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$




So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?



EDIT



To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?






calculus ordinary-differential-equations complex-numbers







share|cite|improve this question











$endgroup$




I have a question stating




Find the complementary function of $frac{d^2y}{dx^2} + 4frac{dy}{dx} + 40y = x$




So I solved the auxillary equation, $ lambda^2 + 4lambda +40 = 0$, which gives $lambda = -2 pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x in mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(Acos{6x} +Bsin{6x})$. To try to get this I did
$$ begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(cos{6x} +isin{6x}) +Be^{-2x}(cos{6x}+isin{6x}) \ &= e^{-2x}((A+B)cos{6x} + (A-B)isin{6x})
end{align}$$
How do you get to the given answer?



EDIT



To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(Acos{6x} +Bsin{6x})$ ?






calculus ordinary-differential-equations complex-numbers




calculus ordinary-differential-equations complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 26 '18 at 19:25







Ewan Miller

















asked Feb 26 '18 at 16:52









Ewan MillerEwan Miller

12713




12713












  • $begingroup$
    You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
    $endgroup$
    – LutzL
    Feb 26 '18 at 17:34


















  • $begingroup$
    You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
    $endgroup$
    – LutzL
    Feb 26 '18 at 17:34
















$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34




$begingroup$
You are aware that all 3 forms are essentially the same, only with different composition of the constants? Is your question on whether the equation $(C,D)=(A+B, (A-B)i)$ is always solvable for $A,B$?
$endgroup$
– LutzL
Feb 26 '18 at 17:34










2 Answers
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$begingroup$

It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    It is easy..



    Use $e^{ix}=cos x+i sin x$.



    And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






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      2 Answers
      2






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      active

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      0












      $begingroup$

      It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$






          share|cite|improve this answer









          $endgroup$



          It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(Ccos{6x}+Dsin{6x})$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 26 '18 at 19:59









          Ewan MillerEwan Miller

          12713




          12713























              -1












              $begingroup$

              It is easy..



              Use $e^{ix}=cos x+i sin x$.



              And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                It is easy..



                Use $e^{ix}=cos x+i sin x$.



                And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  It is easy..



                  Use $e^{ix}=cos x+i sin x$.



                  And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)






                  share|cite|improve this answer











                  $endgroup$



                  It is easy..



                  Use $e^{ix}=cos x+i sin x$.



                  And in final step you can collect $cos$ and $sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 29 '18 at 5:23









                  Siong Thye Goh

                  103k1468119




                  103k1468119










                  answered Dec 29 '18 at 5:00









                  محبس الجنمحبس الجن

                  1




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