How would I find the interval centered about x=0 for which the given IVP has a unique solution: $$(x-2)y''+3y=x$$ With the initial conditions of $y(0)=0$, $y'(0)=1$
The answer is supposedly $(-\infty,2)$. How do I show work for this problem? Or better yet how do I understand it. I believe the DE is homogeneous, and I plugged the DE into Wolfram, and got a complicated output. Any thoughts would be helpful.
My Attempts/Thoughts
To at least attempt, I would say dived the whole thing by $(x-2)$ yielding: $$y''+\frac{3y}{x-2}=\frac{x}{x-2}$$ Making 2 not being a number that can be plugged in...
Hint:
let $$(x-2)y=u(x)$$ So $$y+(x-2)y'=u'$$ $$2y'+(x-2)y''=u''$$ Then $$(x-2)y''+3y=x => $$ $$u''-2y'+3y=x$$ $$u''-2u'/(x-2)+2u/(x-2)^2+3u/(x-2)=x $$ Finally $$(x-2)^2u''-2(x-2)u'+3(x-2)u+2u=x(x-2)^2$$ Now continue...