Partial Fraction Decomposition with Arbitrary Constant in $intfrac{1}{y^4-K^4}dy$.












0












$begingroup$


I need to find the partial decomposition of a fraction which contains an arbitrary constant. This is the final step, or close to final, on a larger non-linear differential equation problem. I need to integrate...




$$intfrac{1}{y^4-K^4}dy$$




Where $K$ is an arbitrary constant. I'm prettry sure I need to break this fraction up with partial fractions. It's been a long while since I've used partial fractions, and I've tried to brush up, but I can't figure out how to split the fraction up into partials with the unknown constant. I know how I would do it if $K$ where an actual known number, and I've tried just going forward with the way that I know, but I'm getting nowhere. How would I go about splitting...



$$frac{1}{y^4-K^4}$$



into partial fractions so that I can integrate it?






integration partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Follow the hint from Herr Graubner and proceed as if $K$ were known.
    $endgroup$
    – Sean Roberson
    Apr 15 '18 at 6:02
















0












$begingroup$


I need to find the partial decomposition of a fraction which contains an arbitrary constant. This is the final step, or close to final, on a larger non-linear differential equation problem. I need to integrate...




$$intfrac{1}{y^4-K^4}dy$$




Where $K$ is an arbitrary constant. I'm prettry sure I need to break this fraction up with partial fractions. It's been a long while since I've used partial fractions, and I've tried to brush up, but I can't figure out how to split the fraction up into partials with the unknown constant. I know how I would do it if $K$ where an actual known number, and I've tried just going forward with the way that I know, but I'm getting nowhere. How would I go about splitting...



$$frac{1}{y^4-K^4}$$



into partial fractions so that I can integrate it?






integration partial-fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Follow the hint from Herr Graubner and proceed as if $K$ were known.
    $endgroup$
    – Sean Roberson
    Apr 15 '18 at 6:02














0












0








0





$begingroup$


I need to find the partial decomposition of a fraction which contains an arbitrary constant. This is the final step, or close to final, on a larger non-linear differential equation problem. I need to integrate...




$$intfrac{1}{y^4-K^4}dy$$




Where $K$ is an arbitrary constant. I'm prettry sure I need to break this fraction up with partial fractions. It's been a long while since I've used partial fractions, and I've tried to brush up, but I can't figure out how to split the fraction up into partials with the unknown constant. I know how I would do it if $K$ where an actual known number, and I've tried just going forward with the way that I know, but I'm getting nowhere. How would I go about splitting...



$$frac{1}{y^4-K^4}$$



into partial fractions so that I can integrate it?






integration partial-fractions







share|cite|improve this question











$endgroup$




I need to find the partial decomposition of a fraction which contains an arbitrary constant. This is the final step, or close to final, on a larger non-linear differential equation problem. I need to integrate...




$$intfrac{1}{y^4-K^4}dy$$




Where $K$ is an arbitrary constant. I'm prettry sure I need to break this fraction up with partial fractions. It's been a long while since I've used partial fractions, and I've tried to brush up, but I can't figure out how to split the fraction up into partials with the unknown constant. I know how I would do it if $K$ where an actual known number, and I've tried just going forward with the way that I know, but I'm getting nowhere. How would I go about splitting...



$$frac{1}{y^4-K^4}$$



into partial fractions so that I can integrate it?






integration partial-fractions




integration partial-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 7 '18 at 6:50









Nosrati

26.5k62354




26.5k62354










asked Apr 15 '18 at 5:56









Phil FernandezPhil Fernandez

413




413












  • $begingroup$
    Follow the hint from Herr Graubner and proceed as if $K$ were known.
    $endgroup$
    – Sean Roberson
    Apr 15 '18 at 6:02


















  • $begingroup$
    Follow the hint from Herr Graubner and proceed as if $K$ were known.
    $endgroup$
    – Sean Roberson
    Apr 15 '18 at 6:02
















$begingroup$
Follow the hint from Herr Graubner and proceed as if $K$ were known.
$endgroup$
– Sean Roberson
Apr 15 '18 at 6:02




$begingroup$
Follow the hint from Herr Graubner and proceed as if $K$ were known.
$endgroup$
– Sean Roberson
Apr 15 '18 at 6:02










2 Answers
2






active

oldest

votes


















2












$begingroup$

Make the ansatz $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$
Multiplying by the denominators we obtain
$$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$
from here you will get an equation system to compute the coefficients






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
    $endgroup$
    – Phil Fernandez
    Apr 16 '18 at 1:31



















0












$begingroup$

$$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



To solve Dr. Graubner's equation



$$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$



as stated above, is a lot of linear algebra. Some of that can be avoided.



$$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



Using $x^4-K^4 = (x-K)(x+K)(x^2+K^2)$, we get



$$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$



Letting $x=K$, you get



begin{align}
1 &= 4AK^3 \
A &= dfrac{1}{4K^3}
end{align}



Letting $x=K$, you get



begin{align}
1 &= -4BK^3 \
B &= -dfrac{1}{4K^3}
end{align}



Letting $x=iK$, you get $1 = -2K^2(D+iCK)$.



Letting $x=-iK$, you get $1 = -2K^2(D-iCK)$.



So $C = 0$ and $D = -dfrac{1}{2K^2}$.



Hence



$$frac{1}{x^4-K^4}=frac{1}{4K^3(x-K)}-frac{1}{4K^3(x+K)}-frac{1}{2K^2(x^2+K^2)}$$



If you don't want to have to resort to complex variables, you can do this after finding the values of $A$ and $B$.



begin{align}
A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \
dfrac{(x+K)(x^2+K^2)}{4K^3}-dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \
dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \
(Cx+D)(x^2-K^2) &= 1 - dfrac{x^2+K^2}{2K^2} \
(Cx+D)(x^2-K^2) &= -dfrac{x^2-K^2}{2K^2} \
Cx+D &= -dfrac{1}{2K^2}
end{align}






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

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    2












    $begingroup$

    Make the ansatz $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$
    Multiplying by the denominators we obtain
    $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$
    from here you will get an equation system to compute the coefficients






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
      $endgroup$
      – Phil Fernandez
      Apr 16 '18 at 1:31
















    2












    $begingroup$

    Make the ansatz $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$
    Multiplying by the denominators we obtain
    $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$
    from here you will get an equation system to compute the coefficients






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
      $endgroup$
      – Phil Fernandez
      Apr 16 '18 at 1:31














    2












    2








    2





    $begingroup$

    Make the ansatz $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$
    Multiplying by the denominators we obtain
    $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$
    from here you will get an equation system to compute the coefficients






    share|cite|improve this answer











    $endgroup$



    Make the ansatz $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$
    Multiplying by the denominators we obtain
    $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$
    from here you will get an equation system to compute the coefficients







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 16 '18 at 13:00

























    answered Apr 15 '18 at 6:01









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    78.4k42867




    78.4k42867












    • $begingroup$
      Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
      $endgroup$
      – Phil Fernandez
      Apr 16 '18 at 1:31


















    • $begingroup$
      Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
      $endgroup$
      – Phil Fernandez
      Apr 16 '18 at 1:31
















    $begingroup$
    Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
    $endgroup$
    – Phil Fernandez
    Apr 16 '18 at 1:31




    $begingroup$
    Should I combine and absorb K into A, B, C, and D during the process of multiplying out? Or should I keep all K as their own entities?
    $endgroup$
    – Phil Fernandez
    Apr 16 '18 at 1:31











    0












    $begingroup$

    $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



    To solve Dr. Graubner's equation



    $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$



    as stated above, is a lot of linear algebra. Some of that can be avoided.



    $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



    Using $x^4-K^4 = (x-K)(x+K)(x^2+K^2)$, we get



    $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$



    Letting $x=K$, you get



    begin{align}
    1 &= 4AK^3 \
    A &= dfrac{1}{4K^3}
    end{align}



    Letting $x=K$, you get



    begin{align}
    1 &= -4BK^3 \
    B &= -dfrac{1}{4K^3}
    end{align}



    Letting $x=iK$, you get $1 = -2K^2(D+iCK)$.



    Letting $x=-iK$, you get $1 = -2K^2(D-iCK)$.



    So $C = 0$ and $D = -dfrac{1}{2K^2}$.



    Hence



    $$frac{1}{x^4-K^4}=frac{1}{4K^3(x-K)}-frac{1}{4K^3(x+K)}-frac{1}{2K^2(x^2+K^2)}$$



    If you don't want to have to resort to complex variables, you can do this after finding the values of $A$ and $B$.



    begin{align}
    A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \
    dfrac{(x+K)(x^2+K^2)}{4K^3}-dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \
    dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \
    (Cx+D)(x^2-K^2) &= 1 - dfrac{x^2+K^2}{2K^2} \
    (Cx+D)(x^2-K^2) &= -dfrac{x^2-K^2}{2K^2} \
    Cx+D &= -dfrac{1}{2K^2}
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



      To solve Dr. Graubner's equation



      $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$



      as stated above, is a lot of linear algebra. Some of that can be avoided.



      $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



      Using $x^4-K^4 = (x-K)(x+K)(x^2+K^2)$, we get



      $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$



      Letting $x=K$, you get



      begin{align}
      1 &= 4AK^3 \
      A &= dfrac{1}{4K^3}
      end{align}



      Letting $x=K$, you get



      begin{align}
      1 &= -4BK^3 \
      B &= -dfrac{1}{4K^3}
      end{align}



      Letting $x=iK$, you get $1 = -2K^2(D+iCK)$.



      Letting $x=-iK$, you get $1 = -2K^2(D-iCK)$.



      So $C = 0$ and $D = -dfrac{1}{2K^2}$.



      Hence



      $$frac{1}{x^4-K^4}=frac{1}{4K^3(x-K)}-frac{1}{4K^3(x+K)}-frac{1}{2K^2(x^2+K^2)}$$



      If you don't want to have to resort to complex variables, you can do this after finding the values of $A$ and $B$.



      begin{align}
      A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \
      dfrac{(x+K)(x^2+K^2)}{4K^3}-dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \
      dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \
      (Cx+D)(x^2-K^2) &= 1 - dfrac{x^2+K^2}{2K^2} \
      (Cx+D)(x^2-K^2) &= -dfrac{x^2-K^2}{2K^2} \
      Cx+D &= -dfrac{1}{2K^2}
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



        To solve Dr. Graubner's equation



        $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$



        as stated above, is a lot of linear algebra. Some of that can be avoided.



        $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



        Using $x^4-K^4 = (x-K)(x+K)(x^2+K^2)$, we get



        $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$



        Letting $x=K$, you get



        begin{align}
        1 &= 4AK^3 \
        A &= dfrac{1}{4K^3}
        end{align}



        Letting $x=K$, you get



        begin{align}
        1 &= -4BK^3 \
        B &= -dfrac{1}{4K^3}
        end{align}



        Letting $x=iK$, you get $1 = -2K^2(D+iCK)$.



        Letting $x=-iK$, you get $1 = -2K^2(D-iCK)$.



        So $C = 0$ and $D = -dfrac{1}{2K^2}$.



        Hence



        $$frac{1}{x^4-K^4}=frac{1}{4K^3(x-K)}-frac{1}{4K^3(x+K)}-frac{1}{2K^2(x^2+K^2)}$$



        If you don't want to have to resort to complex variables, you can do this after finding the values of $A$ and $B$.



        begin{align}
        A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \
        dfrac{(x+K)(x^2+K^2)}{4K^3}-dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \
        dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \
        (Cx+D)(x^2-K^2) &= 1 - dfrac{x^2+K^2}{2K^2} \
        (Cx+D)(x^2-K^2) &= -dfrac{x^2-K^2}{2K^2} \
        Cx+D &= -dfrac{1}{2K^2}
        end{align}






        share|cite|improve this answer









        $endgroup$



        $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



        To solve Dr. Graubner's equation



        $$1=x^3(A+B+C)+x^2(AK-BK+D)+x(AK^2+BK^2-CK^2)+AK^3-BK^3-DK^2$$



        as stated above, is a lot of linear algebra. Some of that can be avoided.



        $$frac{1}{x^4-K^4}=frac{A}{x-K}+frac{B}{x+K}+frac{Cx+D}{x^2+K^2}$$



        Using $x^4-K^4 = (x-K)(x+K)(x^2+K^2)$, we get



        $$1 = A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K)$$



        Letting $x=K$, you get



        begin{align}
        1 &= 4AK^3 \
        A &= dfrac{1}{4K^3}
        end{align}



        Letting $x=K$, you get



        begin{align}
        1 &= -4BK^3 \
        B &= -dfrac{1}{4K^3}
        end{align}



        Letting $x=iK$, you get $1 = -2K^2(D+iCK)$.



        Letting $x=-iK$, you get $1 = -2K^2(D-iCK)$.



        So $C = 0$ and $D = -dfrac{1}{2K^2}$.



        Hence



        $$frac{1}{x^4-K^4}=frac{1}{4K^3(x-K)}-frac{1}{4K^3(x+K)}-frac{1}{2K^2(x^2+K^2)}$$



        If you don't want to have to resort to complex variables, you can do this after finding the values of $A$ and $B$.



        begin{align}
        A(x+K)(x^2+K^2)+B(x-K)(x^2+K^2)+(Cx+D)(x-K)(x+K) &= 1 \
        dfrac{(x+K)(x^2+K^2)}{4K^3}-dfrac{B(x-K)(x^2+K^2)}{4K^3}+(Cx+D)(x-K)(x+K) &= 1 \
        dfrac{x^2+K^2}{2K^2}+(Cx+D)(x^2-K^2) &= 1 \
        (Cx+D)(x^2-K^2) &= 1 - dfrac{x^2+K^2}{2K^2} \
        (Cx+D)(x^2-K^2) &= -dfrac{x^2-K^2}{2K^2} \
        Cx+D &= -dfrac{1}{2K^2}
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 10:36









        steven gregorysteven gregory

        18.3k32358




        18.3k32358






























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            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
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            Redirect URL with Chrome Remote Debugging Android Devices

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            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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