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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...

$$\int\frac{1}{y^4-K^4}dy$$

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?

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

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

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}

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