Let X=x-4:
y"-(3/X)y'+(2/X)2y=X+4 if given "X" should be "x".
Multiply through by X2:
X2y"-3Xy'+4y=X3+4X2.
If one of the solutions to X2y"-3Xy'+4y=0 is y=aXr, (a is a non-zero constant) then:
since y'=dy/dx=dy/dX=arXr-1, y"=d2y/dx2=d2y/dX2=ar(r-1)Xr-2 and:
ar(r-1)Xr-3arXr+4aXr=0=aXr(r(r-1)-3r+4)=aXr(r-2)2, r=2 because Xr≠0 and a≠0.
[NOTE: r(r-1)-3r+4 is the expression arising from Euler's Method. Here it has been derived from scratch.]
y=aX2=a(x-4)2 is a solution of X2y"-3Xy'+4y=0 (that is, y"+(3/(4-x))y'+(2/(x-4))2y=0).
If the original DE was intended to be: y"+(3/(4-x))y'+(2/(x-4))2y=0 then Euler's Method is in fact essentially the solution above.
However, if the DE was intended to be: y"+(3/(4-x))y'+(2/(x-4))2y=x, we still need a solution which involves more than Euler's Method.
If y=y1+y2 is the full solution and y1=a(x-4)2, then y2 (to be found) is a solution to the original DE.
If, though, we start with y"+(3/(4-x))y'+(2/(x-4))2y=X where X=x-4 (not given, but assumed), then let y=aX2+bX3:
y'=2aX+3bX2, y"=2a+6bX
y"-(3/X)y'+(2/X)2y=X, multiply through by X2:
X2y"-3Xy'+4y=X3,
2aX2+6bX3-3X(2aX+3bX2)+4(aX2+bX3)=X3,
2aX2+6bX3-6aX2-9bX3+4aX2+4bX3=X3,
bX3=X3⇒b=1, so y=aX2+X3=X2(a+X)=(x-4)2(a+x-4), making the solution:
y=(x-4)2(x+a-4) where a≠0 is a constant.