I am having trouble understating what this question actually want from me. I am about an hour in too revising this so I am very new, and thought it would be an idea to try and learn it through tacking a problem which is as follows.
Q: Bessel Function of the first kind $J_n(x)$ are solutions of differential equations $$x^2y''+xy'+(x^2-n^2)y=0 \tag{1}$$ use this information to show that $y=xJ_1(x)$ is a solution of $$xy''-y'+xy=0 \tag{2}$$
So my thoughts are from what I have read I have to use the power series solution 'I think' but can seem to see how this gives me $xJ_1(x)$.
Also do I have to rearrange [2] in the form of [1]. My usual method of learning something new is to apply a equation of formula get a general idea of what going on then start to try a build more an intuition and to derive the formula from there.
But I have a feeling I cant quite apply this method to this, could someone maybe give some advice or even a link to a web site. All's that I seem to be getting at the moment are a lot of derivations and no application.
Also I have been told that I have to tick a box to say I am satisfied with my ans. But where is that box?
You have $${{x}^{2}}y''+xy'+\left( {{x}^{2}}-{{n}^{2}} \right)y=0$$ where $$y={{J}_{n}}\left( x \right)$$ Consider now $u=xy$ and the DE $$xu''-u'+xu=0$$ Substituting we have therefore $$x\left( 2y'+xy'' \right)-\left( y+xy' \right)+x\left( xy \right)=0$$ Expanding and collecting $${{x}^{2}}y''+xy'+\left( {{x}^{2}}-1 \right)y=0$$ which is just Bessel’s equation for n=1, hence $u=x{{J}_{1}}\left( x \right)$ satisfies $xu''-u'+xu=0$.