The first-order differential equation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]y' +(ex^2+1+1/x^2)y=0[/tex], with boudary value y(1) =1

Using, asymptotic mathcing to study the behaviour of the sltion as e tends to +0, when x is not too large, the term [tex]ex^2 [/tex] is negligible so an approximate equation for y is

[tex] y'_L +(1+1/x^2)y_L=0 [/tex].

When x is large, [ tex ]ex^2 [ /tex ] is not negligible but [tex]1/x^2 [/tex] is. Therefore, an approximate equation valid as x tends to infinity is [tex] y'_R +(ex^2+1)y_R=0 [/tex].

I think that te upper edge of the left region would be the largest value of x for which ex^2 is still small compared with 1. This would suggest that the left region consists of those x for which x<<e^-(1/2) as e tends to +0. But actually the region of validity of the left solution is e^(-1/3) (e tends to +0)....Can anyone explain this to me??

Sams as the right region...Please kindly help

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# Asymptotic mathcing for a first order differential equation

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