Approximate solution to DE - confused

1. Oct 11, 2011

1. The problem statement, all variables and given/known data
This is a problem from an old exam I am reviewing for practice.

Find a good approximation, for x large and positive to the solution of the following equation:

y''-(3/x)y'+(15/(4x^2)+x^(1/2))y=0

Hint: remove first derivative term
2. Relevant equations

3. The attempt at a solution I'm not sure first of all that I understand the logic behind the hint. After taking this suggestion, my first inclination is to throw away the 15/4x^2 term, leaving y''+sqrt(x)y=0. This is an equation that while it seems simple, I'm ashamed to say I can't figure out how to solve.

Also, according to the problem's solution, you actually throw away just the x^(1/2)y term. This leaves an easily solvable homogeneous Cauchy-Euler DE. But the logic doesn't seem to make sense to me. The solution says that x^(1/2)y and (3/x)y' are negligible at large x. However, isn't the 15/4x^2 term much smaller than these terms?

2. Oct 11, 2011

Ray Vickson

Would it help to set t=1/x, then look for a small-t approximation? I haven't tried it and have no access to a computer at the moment.

RGV

3. Oct 11, 2011

jackmell

Tell you what, how might you confirm your suspicions? I mean what would you do to figure this out on your own? What about just running the DEs and checking for yourself? How about start with x0=10 000 say for y(x0)=1 and y'(x0)=1. That's large enought huh? Now, run them numerically, first the original DE, then run versions of the DE without one or more of the other terms and then see how well the simplified versions approximate the original equation. Here's what I would do in Mathematica. At least at x=10,000, it's pretty convincing to me what's important and not important to leave in for large x.

Code (Text):

x0 = 10000;
y0 = 1;
y1 = 1;

mysol = NDSolve[{Derivative[2][y][x] - (3/x)*Derivative[1][y][x] + (15/(4*x^2) + Sqrt[x])*y[x] == 0, y[x0] == y0,
Derivative[1][y][x0] == y1}, y, {x, x0, x0 + 10}]

p1 = Plot[y[x] /. mysol, {x, x0, x0 + 10}]

mysol2 = NDSolve[{Derivative[2][y][x] - (3/x)*Derivative[1][y][x] + (15/(4*x^2))*y[x] == 0, y[x0] == y0,
Derivative[1][y][x0] == y1}, y, {x, x0, x0 + 10}]

p2 = Plot[y[x] /. mysol2, {x, x0, x0 + 10}]

mysol3 = NDSolve[{Derivative[2][y][x] + Sqrt[x]*y[x] == 0, y[x0] == y0, Derivative[1][y][x0] == y1}, y,
{x, x0, x0 + 10}]

p3 = Plot[y[x] /. mysol3, {x, x0, x0 + 10}]

Show[{p1, p3}]

4. Oct 11, 2011

I'm not sure I follow. I'll keep looking, but I'm somewhat new to Mathematica.

5. Oct 11, 2011

Ah, so I am correct, and it is a mistake in the solution (which was actually written by a student) ? I ran it in mathematica, and it appears to be so. I should be able to solve my approximate equation by hand but can't think of how. Any suggestions? Thanks so much for taking time to look at this.

6. Oct 12, 2011

jackmell

I'm not entirely sure about all this but seems to me based on the numerical results that for large x, we can neglect the terms with x in the denominator and we get as a good approximation:

$$y''+\sqrt{x}y=0$$

. . . well that ain't easy neither but I think we can solve it manually by making a change of independent variable by letting $u=x^{5/4}$. I ran through this substitution quick and got it close to the solution reported by Mathematica in terms of Bessel functions. I'm basing this substitution on the results reported by Mathematica for the code:

DSolve[y''[x]+x^(1/n) y[x]==0,y,x]