- #1

Ryker

- 1,086

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## Homework Statement

Find all solutions of [tex](x - 2)(x - 3)y’ + 2y = (x - 1)(x - 2)[/tex] on each of the following intervals: [tex](a) (-\infty, 2); (b) (2, 3); (c) (3, +\infty).[/tex]

## 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

I chose a = 1, first concentrating on a point in interval (a). Using that, I arrived to the solution [tex]f(x) = 4b (\frac{x-2}{x-3})^{2} + \frac{(x-2)(x-1)^{2}}{(x-3)^{2}},[/tex]which is basically the same as what WolframAlpha gives you, namely [tex]f(x) = c_{1} e^{-2(\log(3-x)-\log(2-x))} + \frac{(x-2)(x-1)^{2}}{(x-3)^{2}}.[/tex]The only difference I see is that their solution is general, and Apostol seems to not use the integration constant, but rather operates with b, as in f(a) = b. I'd say that's just a matter of style, though.

What I don't see here is why and how there would be different solutions for the different intervals (a), (b) and (c). As long as the function is defined there, shouldn't they all be the same? And the function is defined there, I think, it's just that WolframAlpha forgot to put absolute values in natural logs. Also, for a, plugging in values from intervals (b) and (c) just gives you a different constant, but that's it, so I'm really puzzled by this one.

Thanks in advance for any help.

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