Question

Solving this following ODE by Frobenius Method
X^2y’’+(4x-x^2)y’+(2-x)y=0

Answer 1

$$x^2y''+(4x-x^2)y'+(2-x)y=0 \ \ \ \ \ \ \ \ \ (1)$$

Dividing by $x^2$

$$y''+{4x-x^2 \over x^2}y'+{2-x\over x^2}y=0$$ $$xP(x)=4-x, x^2Q(x)=2-x$$

Both $P(x)$ and $Q(x)$ are analytic at $x=0.$ A singular point $x=0$ is a regular singular point of the equation (1). Then we can find at least one solution to a second order differential equation by assuming a solution of the form: 

$$y=\displaystyle\sum_{n=0}^\infin a_nx^{n+r}, a_0\not =0$$ $$y'=\displaystyle\sum_{n=0}^\infin a_n(n+r)x^{n+r-1}$$ $$y''=\displaystyle\sum_{n=0}^\infin a_n(n+r)(n+r-1)x^{n+r-2}$$

Substitute

$$x^2 \displaystyle\sum_{n=0}^\infin a_n(n+r)(n+r-1)x^{n+r-2}+(4x-x^2)\displaystyle\sum_{n=0}^\infin a_n(n+r)x^{n+r-1}+$$ $$+(2-x)\displaystyle\sum_{n=0}^\infin a_nx^{n+r}=0$$

Or

$$\displaystyle\sum_{n=0}^\infin a_n(n+r)(n+r-1)x^{n+r}+\displaystyle\sum_{n=0}^\infin a_n[4(n+r)+2]x^{n+r}+$$ $$+\displaystyle\sum_{n=1}^\infin a_{n-1}(-n-r)x^{n+r}=0$$

$$a_0r(r-1)x^r+a_0(4r+2)x^r=0$$ $$(r+2)(r+1)=0$$ $$r_1=-2, r_2=-1$$

$$a_1(1+r)rx^{r+1}+a_1(4(1+r)+2)x^{r+1}+a_0(-1-r)x^{1+r}=0$$ $$a_1(r+2)(r+3)x^{r+1}=a_0(1+r)x^{1+r}$$

$$a_2(2+r)(1+r)x^{r+2}+a_2(4(2+r)+2)x^{r+2}+a_1(-2-r)x^{r+2}=0$$ $$a_2(r+3)(r+4)x^{r+2}=a_1(2+r)x^{2+r}$$

$$a_3(3+r)(2+r)x^{r+3}+a_3(4(3+r)+2)x^{r+3}+a_2(-3-r)x^{r+3}=0$$ $$a_3(r+4)(r+5)x^{r+3}=a_2(3+r)x^{3+r}$$

$$a_n(r+n+1)(r+n+2)x^{r+n}=a_{n-1}(n+r)x^{n+r}$$

$$r_1=-2$$ $$a_1(-2+2)(-2+3)x^{-2+1}=a_0(1-2)x^{1-2}, a_0=0$$

$$r_2=-1$$

$$a_1(-1+2)(-1+3)x^{-1+1}=a_0(1-1)x^{1-1}, a_1=0, a_0\not =0$$

$$a_2(-1+3)(-1+4)x^{-1+2}=a_1(-1+r)x^{-1+r}, a_2=0$$

$$a_2=a_3=...=a_n=...=0$$

$$\widetilde{y_2}=a_0x^{0-1}={a_0 \over x}$$

$$y={a_0 \over x}, a_0\in \R, a_0\not= 0$$