How to Obtain Fundamental Solutions for Non-Constant Coefficient Equations?

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Solve by method of variation of parameters
(x^2)y'' - (4x)y' + 6y = x^4*sinx (x > 0)

Hey, I know how to solve problems using variation of parameters but only when the corresponding homogenous equation has constant coefficients...

y'' - (4/x)y' + (6/x^2)y = 0.. the bit I am confused about is how to obtain the fundamental solutions to this equation {y1, y2} when the coefficients are not constants. Any help would be appreciated.

Thanks.
 
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Note that each term has the same "units", if you think of x as having units of length (so that a derivative removes one unit of length). Such differential equations have solutions of the form x^r. Plug this in and solve for r, and you'll quickly see why such solutions work. This is something you can just remember, although it would also fall out if you tried the Frobenius method.
 
im not quite sure I understand where to plug in x^r .
 
I mean y(x)=xr is a solution to the homogenous differential equation for certain r. Plug in this y and see which r work.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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