So i was told to find the general solution $$2x^2y''-xy'+(x-5)y=0$$ (1) using power series. After a little manipulation (1) becomes $$y''-\frac{1}{2x}y'+\frac{(x-5)}{2x^2}y=0$$ $$P(x)=\frac{-1}{2x}$$ $$Q(x)=\frac{x-5}{2x^2}$$ $$p(x)=xP(x)=\frac{-1}{2}$$ $$q(x)=x^2Q(x)=\frac{x-5}{2}$$ $$p(0)=\frac{-1}{2}$$ $$q(0)=\frac{-5}{2}$$ thus the indicial equation is $$r(r-1)-\frac{1}{2}r-\frac{5}{2}=0$$ and therefore $$2r^2-3r-5=0$$ $$r_1=-1$$ and $$r_2=\frac{5}{2}$$. So for $$r=-1$$ $$y=\sum_{n=0}^\infty a_nx^{n-1}$$ $$y'=\sum_{n=0}^\infty(n-1)a_nx^{n-2}$$ and $$y''=\sum_{n=0}^\infty(n-1)(n-2)a_nx^{n-3}$$ And now the differential equation $$2x^2y''-xy'+(x-5)y=0$$ becomes $$2x^2\sum_{n=0}^\infty(n-1)(n-2)a_nx^{n-3}-x\sum_{n=0}^\infty(n-1)a_nx^{n-2}+x\sum_{n=0}^\infty a_nx^{n-1}-5\sum_{n=0}^\infty a_nx^{n-1}=0$$
thus $$\sum_{n=0}^\infty2(n-1)(n-2)a_nx^{n-1}-\sum_{n=0}^\infty(n-1)a_nx^{n-1}+\sum_{n=0}^\infty a_nx^n-\sum_{n=0}^\infty 5a_nx^{n-1}=0$$
For the term $$\sum_{n=0}^\infty a_nx^n$$ i replaced $n$ with $n-1$ in order for the powers of x to match and therefore we now have $$\sum_{n=1}^\infty a_{n-1}x^{n-1}$$ Substituting back to the equation i end up with $$\sum_{n=0}^\infty2(n-1)(n-2)a_nx^{n-1}-\sum_{n=0}^\infty(n-1)a_nx^{n-1}+\sum_{n=1}^\infty a_{n-1}x^{n-1}-\sum_{n=0}^\infty 5a_nx^{n-1}=0$$
For $n=0$ -> $$5a_0x^{-1}-5a_0x^{-1}=0$$ so $n$ has to be 1 and over. Therefore $$\sum_{n=0}^\infty [2(n-1)(n-2)a_n-(n-1)a_n+a_{n-1}-5a_n]x^{n-1}=0$$ thus $$a_n=\frac{-a_{n-1}}{n(2n-7)}$$ Here's where i am stuck and i really need your help.. i try to find the pattern by substituting $n=1$ ,$n=2$ etc and for $n=1$ i get $a_1=\frac{a_0}{5}$ for $n=2$ i get $a_2=\frac{1a_0}{5!}$ and for $n=3$ i get $a_3=\frac{1}{3}\frac{1a_0}{5!}$ Am i doing something wrong? Thanks in advance for any idea or possible solution!