- #1
pretendinitis
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Homework Statement
The question asks me to find the general solution to [tex](x^2-1)y' + 2y = (x+1)^2[/tex], 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 [tex]y' + \frac{2}{x^2 - 1}y = \frac{(x + 1)^2}{x^2-1}[/tex], with [tex]P(x)=\frac{2}{x^2 - 1}[/tex] and [tex]f(x)=\frac{(x + 1)^2}{x^2-1}[/tex]. The integrating factor is [tex]e^{\int2/(x^2 - 1) dx} = e^\ln|x-1|-\ln|x+1|}=\frac{x-1}{x+1}[/tex].
Substituting back into the original equation:
[tex]\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}[/tex]
[tex]\left(\frac{x-1}{x+1}y\right)'=1[/tex]
Integrating:
[tex]\int{\left(\frac{x-1}{x+1}y\right)'} dx=\int{1} dx[/tex]
[tex]\frac{x-1}{x+1}y=x+c[/tex]
[tex](x-1)y=x(x+1)+c(x+1)[/tex]
[tex]y=\frac{x+1}{x-1}(x+c)[/tex]
The Attempt at a Solution
I know that [tex]P(x)[/tex] and [tex]f(x)[/tex] are both discontinuous at -1 and 1. But then, what is the largest interval over which the general solution is defined? Is it [tex]-\infty < x < -1[/tex], [tex]-1 < x < 1[/tex], or [tex]1 < x <\infty[/tex]? I feel like there's something really simple here that I'm missing. Any help is appreciated.