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
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    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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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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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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