Solving Homework: Finding General Solution & Largest Interval

[pretendinitis]

Homework Statement

The question asks me to find the general solution to $$(x^2-1)y' + 2y = (x+1)^2$$, and to determine the largest interval over which this general solution is defined. It's the latter which is tripping me up.

Homework Equations

I've already found the general solution as follows: rearranging the above equation gives $$y' + \frac{2}{x^2 - 1}y = \frac{(x + 1)^2}{x^2-1}$$, with $$P(x)=\frac{2}{x^2 - 1}$$ and $$f(x)=\frac{(x + 1)^2}{x^2-1}$$. The integrating factor is $$e^{\int2/(x^2 - 1) dx} = e^\ln|x-1|-\ln|x+1|}=\frac{x-1}{x+1}$$.

Substituting back into the original equation:
$$\frac{x-1}{x+1}y'+\frac{x-1}{x+1}\frac{2}{x^2 - 1}y=\frac{x-1}{x+1}\frac{x+1}{x-1}$$

$$\left(\frac{x-1}{x+1}y\right)'=1$$

Integrating:
$$\int{\left(\frac{x-1}{x+1}y\right)'} dx=\int{1} dx$$

$$\frac{x-1}{x+1}y=x+c$$

$$(x-1)y=x(x+1)+c(x+1)$$

$$y=\frac{x+1}{x-1}(x+c)$$

The Attempt at a Solution

I know that $$P(x)$$ and $$f(x)$$ are both discontinuous at -1 and 1. But then, what is the largest interval over which the general solution is defined? Is it $$-\infty < x < -1$$, $$-1 < x < 1$$, or $$1 < x <\infty$$? I feel like there's something really simple here that I'm missing. Any help is appreciated.

 

[Mark44]

I don't think there is a problem at x = -1. If I'm wrong, I hope someone will jump in and correct my thinking.

The solution you found is defined for x = -1. In that case y = 0, so y' = 0. This function satisfies your original differential equation, so it looks to me like your solution is defined on two intervals:
$$-\infty < x < 1 \cup 1 < x < \infty$$

They are both the same size, meaning both intervals have the same number of numbers in them, so for the largest interval, you could pick either one.

 

[pretendinitis]

Thanks Mark. My textbook for this class, Advanced Engineering Mathematics (3rd edition) by Zill and Cullen, implies that I should choose an interval for the general solution which $$P(x)$$ and $$f(x)$$ are both continuous on...

EDIT: I've just noticed the unsimplified form for my integrating factor, $$e^{\ln|x-1|-\ln|x+1|}$$, appears to be only valid for $$x > 1$$, which would make $$1 < x <\infty$$ the only interval allowed. Is that right?

 

[Mark44]

Works for me.