Re: Diff. Eq. Series Problem (URGENT)

 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.

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?