# Asymptotic mathcing for a first order differential equation

1. Jul 31, 2006

### hanson

The first-order differential equation
$$y' +(ex^2+1+1/x^2)y=0$$, 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 $$ex^2$$ is negligible so an approximate equation for y is
$$y'_L +(1+1/x^2)y_L=0$$.

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

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

Last edited: Aug 1, 2006
2. Jul 31, 2006

### Staff: Mentor

I'm confused - $$e$$ is usually a constant
$$e~=2.71828182845904523536028747135266249775724709369995...$$

3. Jul 31, 2006

### hanson

sorry, I am referring e to be a small pertubation introduced...
It would better to use a Greek word for it...
but I have forgotten the word...

4. Jul 31, 2006

### HallsofIvy

e= epsilon: $\epsilon$

5. Jul 31, 2006

### hanson

yes, HallsofIvy! Thank you.

Do you mind helping me for this question?