- #1

Ryker

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## Homework Statement

Alright, so I started studying ODE's on my own, and am using Apostol's Calculus to that regard. Very quickly I've stumbled upon a problem I can't quite get my head around, namely problem 7 in Chapter 8.3. So here it is:

"Find all solutions of x(x + 1)y’ + y = x(x + 1)

^{2}e

^{-x2}on the interval (-1, 0). Prove that all solutions approach 0 as x tends to - 1, but that only one of them has a finite limit as x tends to 0.

## Homework Equations

[tex]f(x) = be^{-A(x)} + e^{-A(x)} \int_a^x \! Q(t)e^{A(t)}dt[/tex]

## The Attempt at a Solution

Since we have to find solutions on the interval (-1, 0), I chose a = -1/2. This is where I might have gone wrong, but in an example Apostol just arbitrarily chooses a point in the interval you want and goes with it to find solutions. So should I have chosen a different a, and if yes, why? I couldn't have chosen 0 or -1, since ln0 is undefined, and so is x/(x+1) for x = -1.

Anyway, this is what I got for P(x), A(x), Q(x) and f(x).

[tex]P(x) = \frac{1}{x(x+1)}[/tex]

[tex]A(x) = \int_a^x \! P(t)dt = \int_\frac{-1}{2}^x \! P(t)dt = \ln (-\frac{x}{x+1})[/tex]

[tex]Q(x) = (x+1)e^{-x^2}[/tex]

So with a = -1/2, [tex]f(x) = \frac{-b(x+1)}{x} - \frac{1}{2}(e^{-x^2} - e^{\frac{-1}{4}})[/tex]

With x tends to -1, the series expansion of e

^{-x2}is [tex]e^{-x^2} = \frac{1}{e} + \frac{2(x+1)}{e} + o(x+1),[/tex] which when plugging into f(x) doesn't give you that all solutions to 0, but rather to [tex]\frac{1}{e} - \frac{1}{\sqrt[4]{e}}.[/tex]Any help?

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