Finding the Form of L: An Example

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SUMMARY

The discussion focuses on solving the differential equation represented by the operator L, specifically in the form x^2(y'') - 2x(y') + 2y = x^3ln(x). The method involves determining the kernel by solving Ly = 0, leading to the suggested trial solution y = x^m. By substituting this form into the equation, the characteristic equation m^2 - 3m + 2 = 0 is derived, yielding roots m = 1 and m = 2, which results in the general solution y = Cx + Dx^2. The key takeaway is understanding how to derive the suggested form for L from the associated homogeneous equation.

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An example in my notes.

To solve for green function of a differential equation.
eg x^2(y'') - 2x(y') + 2y = x^3ln(x)

The first thing is to solve the kernel by solving Ly = 0
and it says the form of L suggests trying y = x^m

How to I obtain the suggested form of L?
 
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By taking the suggestion?
You are told to consider the equation Ly= 0, the "associated homogeneous" equation to your non-homgeneous equation. Since you are given the equation x2y"- 2xy'+ 2y= x3ln x, it follows that Lx= x2y"- 2xy'+ 2y= 0.


If y= xm, then y'= mxm-1 and y"= m(m-1)xm-2. Putting that into the equation, Lx= 0, you get
x2(m(m-1)xm-2- 2x(mxm-1)+ 2xm=- m(m-1)xm- 2m xm+ 2xm= 0 so either x=0 or m(m-1)- 2m+ 2= m2- 3m+ 2= (m-2)(m-1)= 0. m= 1 or m= 2 which gives you the general solution y= Cx+ Dx2.
 
Thx... I understand how to get the general solution but how to I come up with a suggested form?

Say i was given a differential equation. For example Ly = y'' - alpha^2*y = h(x). So first i need to make the equation to Ly = 0. But what is the suggested form?
 

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