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The function $w(x)$ satisfies $$\frac{d^2w}{dx^2} + 2x\frac{dw}{dx} - 2\beta w = 0$$ with $$w(\infty)=0 \text{ and } 0<\beta < 1.$$ By writing $w(x)= exp(-x^2)f(x)$, obtain the first two terms of an asymptotic expansion for $w(x)$ as $x \to \infty $.

smith
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  • And what have you done so far? – Clement C. Oct 21 '15 at 18:30
  • I have differentiated $w(x)$ and plugged it back into the equation .... then found that $$2\beta f(x) + 2x \frac{df}{dx} + 2f(x) - \frac{d^2f}{dx^2}=0$$ Then I took the leading terms which I think are $2x \frac{df}{dx} + 2f(x) \sim 0$

    I'm currently trying to solve this? not too sure if any of this is right though

     – smith Oct 21 '15 at 18:31
  • I'm assuming what I done is incorrect because I get $$f(x) = \frac{c_1}{x}$$ – smith Oct 21 '15 at 18:39
  • I get the same differential equation on $f$ (at first glance), but I don't think it has a nice closed-form solution like the one you suggest. – Clement C. Oct 21 '15 at 18:54
  • surely it works because if I substitute my $f(x)$ into my approximation I get $$\frac{2C_1}{x} + 2x\frac{-c_1}{x^2} \sim 0 $$ $$\Rightarrow \frac{2C_1}{x} - \frac{2C_1}{x} =0 $$ – smith Oct 21 '15 at 18:58
  • @smith There is an approach that can be used to recover the entire asymptotic series. If you repost and let me know, I am happy to help. – Mark Viola Oct 22 '15 at 15:58
  • Thank you @Dr.MV I shall repost it now, this question has been bugging me for hours now – smith Oct 22 '15 at 16:24
  • @Dr.MV http://math.stackexchange.com/questions/1492551/asymptotic-expansion-from-3-leading-terms – smith Oct 22 '15 at 16:33

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