Solving Problem 7 in Apostol's Calculus Ch. 8.3

[Ryker]

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)²e^-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

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

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).

P(x) = \frac{1}{x(x+1)}
A(x) = \int_a^x \! P(t)dt = \int_\frac{-1}{2}^x \! P(t)dt = \ln (-\frac{x}{x+1})
Q(x) = (x+1)e^{-x^2}

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

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

 

[hunt_mat]

This is an integrating factor problem, you need to calculate:
<br /> p(x)=\int\frac{dx}{x(x+1)}<br />
and then compute exp(p(x))

 

[Ryker]

Yeah, this is what I did with A(x), but according to Apostol A(x) is an definite integral, not an indefinite one. So why would I need to consider the latter?

 

[hunt_mat]

I arrive at:
<br /> \frac{d}{dx}\left(\frac{xy}{1+x}\right) =\frac{2xe^{-x^{2}}}{1+x}<br />

 

[Ryker]

Damn, you lost me. Why exactly did you do that, and what did you use for y?

 

[hunt_mat]

Okay, divide through by x(1+x) to obtain:
<br /> y&#039;+\frac{y}{x(1+x)}=2e^{-x^{2}}<br />
Compute the integrating factor to obtain:
<br /> \frac{x}{x+1}<br />
Multiplying through by this shows what I wrote.

 

[Ryker]

Oh, sorry, I forgot to superscribe 2 next to (x + 1) in the original equation, and have now edited the original post to do that. So the equation is actually x(x + 1)y’ + y = x(x + 1)²e^-x2. Sorry for the confusion. Any ideas now with the proper equation?

 

[hunt_mat]

Okay then you would have:
<br /> \frac{d}{dx}\left(\frac{xy}{1+x}\right) =2xe^{-x^{2}}<br />
Integrate this to obtain:
<br /> y=C\frac{1+x}{x}-e^{-x^{2}}<br />
Where C is an integration constant.

 

[Ryker]

How do you get that first part? Being just done with first year, I might be missing knowledge that would help me see it.

Also, if you arrive to the y you have, and then approximate e^-x2 with a Taylor polynomial (you probably don't even have to here), you get that as x tends to -1, y tends to -1/e, not 0. Again, am I missing something here?

 

[hunt_mat]

Do you mean the left hand side?

 

[Ryker]

Both, I think

 

[hunt_mat]

Right:
<br /> y&#039;+\frac{y}{x(1+x)}=2(1+x)e^{-x^{2}}<br />
Multiply by the integrating factor x/(1+x) to obtain:
<br /> \frac{xy&#039;}{1+x}+\frac{y}{(1+x)^{2}}=2xe^{-x^{2}}<br />
Does that help?

 

[Ryker]

Yeah, I see how you got that now, except that I think there shouldn't be a 2 on the right hand side, but rather just e^-x2. There's namely no 2 in the original equation, so I'm guessing this is just a remnant of my original post, where I forgot to superscribe that 2 you're now using. It probably doesn't matter for our purposes, but just wanted to point that out to see whether there's something else I'm missing.

In any case, this seems to be what I got using Apostol's way, as well, it's just that I followed a previous example of his and did definite integration with "a" randomly chosen from the interval. Why would that be wrong, though?

OK, so now we have the equation for f(x), but I still don't see how all solutions tend to 0 as x tends to -1. Taking the limit of f(x) this way namely results in 1/e (or 1/2e if you don't have that 2).

 

[Ryker]

Any further comments? I'm namely still unable to get to the bottom of this.

edit: Nevermind, actually. I found my mistake

Using f(x) = be^{-A(x)} + e^{-A(x)} \int_a^x \! Q(t)e^{A(t)}dt, I said I arrived to f(x) = \frac{-b(x+1)}{x} - \frac{1}{2}(e^{-x^2} - e^{\frac{-1}{4}}), and clearly I omitted the e^{-A(x)} part somewhere along the line.

So the correct answer for f(x) would be: f(x) = \frac{-b(x+1)}{x} - \frac{x + 1}{2x}(e^{-x^2} - e^{\frac{-1}{4}}), which when x\to -1 clearly approaches 0.

Thanks for the help, hunt_mat!

 

[hunt_mat]

That's why I come here, to see fun maths questions and to help people.