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

Question

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?

Answer 1

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

Answer 2

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

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}

Answer 3

$$ \begin{aligned} \therefore \frac{1}{y^{4}-K^{4}} &=\frac{1}{\left(y^{2}+K^{2})\left(y^{2}-K^{2}\right)\right.} \\ &=\frac{1}{2 K^{2}}\left(\frac{1}{y^{2}-K^{2}}-\frac{1}{y^{2}+K^{2}}\right) \\ &=\frac{1}{2 K^{2}}\left[\frac{1}{2 K}\left(\frac{1}{y-K}-\frac{1}{y+K}\right)-\frac{1}{y^{2}+K^{2}}\right] \\ &=\frac{1}{4 K^{3}(y-K)}-\frac{1}{4 K^{3}(y+K)}-\frac{1}{2 K^{2}\left(y^{2}+K^{2}\right)} \\ \therefore \quad \int \frac{1}{y^{4}-K^{4}} d y &=\frac{1}{4 K^{3} }\ln \left|\frac{y-K}{y+K}\right|-\frac{1}{2 K^{3}} \tan ^{-1}\left(\frac{y}{K}\right)+C \end{aligned} $$