Question

integrate dy/dx = (y^2-1)/(x^2 - 1)
my solutions manual says that simplifies to ....
(1/2)(1/(y-1) - 1/(y+1)) dy= (1/2)(1/(x-1) - 1/(x+1)) dx
how do they reach this answer?

Answer 1

\(\large\color{slate}{ dy/dx = (y^2-1)/(x^2 - 1)}\)
dividing both sides by y^2-1, and multiplying times dx on both sides:
\(\large\color{slate}{ dy/(y^2-1) =dx/(x^2-1) }\)

then both sides go by partical fraction.

Answer 2

Partial fractions

Answer 3

That's the short answer. Here's a more detailed explanation: The goal is to be able to write a rational function \(R(x)=\dfrac{P(x)}{Q(x)}\) as a sum of rational functions made up of the irreducible factors of the denominator, \(Q(x)\).

In this case, you have
\[R(x)=\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}\]
We have two linear factors, so we suppose we can split up \(R(x)\) into the following:
\[\frac{1}{(x-1)(x+1)}=\frac{a_1}{x-1}+\frac{a_2}{x+1}\]
for some real \(a_1,a_2\). If we distribute the denominator on the LHS to the RHS and cancel the relevant factors, we have
\[1=a_1(x+1)+a_2(x-1)~~\iff~~0x+1=(a_1+a_2)x+(a_1-a_2)\]
Matching up coefficients for corresponding powers of \(x\), we have
\[\begin{cases}a_1+a_2=0\\a_1-a_2=1\end{cases}\]

Answer 4

Thank you so much!