Diff. Eq. Series Problem (URGENT)

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Randy - 27 Apr 2006 04:23 GMT

Every once in a while my professor makes the homework from scratch, i.e. not using problems from the book. While I really hate this style because I can't check my work, I have an assignment due tomorrow and I'm completely stumped. Here it is:

Find the general solution of

2(x^2)y'' + 3xy' - y = 0, x>0

using the fact that y1(x) = 1/x is a solution.

I've got no examples of this in my textbook, or any online help thread that I can find. I have 4 problems like this on this assignment; help as soon as possible please?

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Paul Sperry - 27 Apr 2006 06:09 GMT

In article
<11951639.1146108282575.JavaMail.jakarta@nitrogen.mathforum.org>, Randy
<rmheubel@syr.edu> wrote:

> Every once in a while my professor makes the homework from scratch, i.e. not
> using problems from the book. While I really hate this style because I can't
[quoted text clipped - 9 lines]
> I've got no examples of this in my textbook, or any online help thread that I
> can find. I have 4 problems like this on this assignment; help as soon as possible please?

Rewrite as y'' + p(x)y' + q(x)y = 0

y_2 = y_1*int((1/y_1^2)exp(-int(p(x),x),x).

This must be in your text somewhere.

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Paul Sperry
Columbia, SC (USA)

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Randy - 27 Apr 2006 06:27 GMT

I promise you, I've searched not only the section we're supposed to be in but the entire chapter on series equations, and nothing involving a given fundamental equation.

That being said, can you elaborate a little on that? And maybe make it easier to read?

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Paul Sperry - 27 Apr 2006 06:56 GMT

In article
<2227025.1146115693729.JavaMail.jakarta@nitrogen.mathforum.org>, Randy
<rmheubel@syr.edu> wrote:

> I promise you, I've searched not only the section we're supposed to be in but
> the entire chapter on series equations, and nothing involving a given
> fundamental equation.
>
> That being said, can you elaborate a little on that? And maybe make it easier
> to read?

It is a standard result for second order homogeneous DEs with
non-constant coefficients and one known soulution. No series is
involved.

I can't do much about the notation - write it out with the integral
sign replacing the "int" and be careful with parentheses.

If you insist on a series solution, try Google. In ASCII, for this
newsgroup, it would be an ungodly and unreadable mess.

[By the way, y_2 = (2/3)x^(1/2).]

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Paul Sperry
Columbia, SC (USA)

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G.E. Ivey - 27 Apr 2006 21:17 GMT

> Every once in a while my professor makes the homework
> from scratch, i.e. not using problems from the book.
[quoted text clipped - 12 lines]
> like this on this assignment; help as soon as
> possible please?

You know, I'm sure, that if you know one solution, a, to a polynomial equation, you can divide by (x-a)to decrease the degree of the equation. There is a similar technique for differential equations.
If a(x) is a solution to a differential equation then replacing y with a(x)u(x) will give you a differential equation of lower degree for u(x).

In this case, replace y with u(x)/x. Then y'= (xu'- u)/x^2 and y"= ((u'+ xu"- u')x^2- 2x(xu'-u))/x^4= (x^3u"- 2x^2u'+ 2xu)/x^4.

Then 2x^2y"= (2x^3u"- 4x^2u'+ 4xu)/x^2= 2xu"- 4u'+ 4u/x and 3xy'= 3xu'- u)/x= 3u'- u/x so
2x^2y"+ 3xy'- y= 2xu"- 4u'+ 4u/x+ 3xu'-3u/x- u/x=
2xu"- u'= 0. Do you see how the "u" terms canceled. Each time you use the product rule (or quotient rule) there is always a term where you have not differentiated the "u" at all. Since the term you have differentiated in those cases (here 1/x) satisfies the equation, it all cancels out. Now, let v= u' so the equation is
2xv'- v= 0. That's an easy separable equation for v. Once you've integrated that to find u, multiply by 1/x to find a second, independent, solution to the differential equation. The result will be the same as that formula the other response gave you.

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G.E. Ivey - 28 Apr 2006 18:40 GMT

Are you aware that you can always check your solution by substituting it into the equation and seeing if it works?

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Math Forum / Mathematics / Undergraduate Math / April 2006

Diff. Eq. Series Problem (URGENT)