Finding the solution to $xy'' +2y' +xy=0$ around $x_{0}=0$using the method of Frobenius.

Question

We know that the solution of this ODE is like: $$ y=\sum_{n=0}^{\infty}C_nx^{n+r}$$

Them derivative $y$ and $y'$. $$y'=\sum_{n=0}^{\infty}(n+r)C_nx^{n+r-1}$$ $$y''=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_nx^{n+r-2}$$

Replace $y$ , $y'$ and $y''$ in the ODE. $$\sum_{n=0}^{\infty}(n+r-1)(n+r)C_nx^{n+r-1}+\sum_{n=0}^{\infty}2(n+r)C_nx^{n+r-1} +\sum_{n=0}^{\infty}C_nx^{n+r+1}=0 $$ Now in the first and second summations if $k=n-1$ and in the third summation if $k=n+1$. We have:

$$\sum_{k=-1}^{\infty}(k+r)(k+r+1)C_{k+1}x^{k+r}+\sum_{n=-1}^{\infty}2(k+r+1)C_{k+1}x^{k+r} +\sum_{n=1}^{\infty}C_{k-1}x^{k+r}=0 $$ Now I separed the $k=-1$ and $k=0$ in the first and second summantions. $$r(r-1)C_0x^{r-1}+(r+1)(r)C_1x^{r} + 2rC_0x^{r-1} +2(r+1)C_1x^r + \sum_{k=1}^{\infty}[(k+r)(k+r+1)C_{k+1}x^{k+r}+2(k+r+1)C_{k+1}x^{k+r}+C_{k-1}x^{k+r}]=0$$ After that I equalized the right side with the left and I have this 2 equation to find $r$: $$(r^2 +r)C_0=0$$ and $$(r^2+3r+2)C_{1}=0$$ If you solve the quadratic equations you have: $$r_1=0$$ $$r_{2,3}=-1$$ $$r_4=-2$$ And that's all that I did please help me.

Answer 1

$$xy'' +2y' +xy=0\qquad ......(1)$$

$~x=0~$ is a regular singular point of equation $(1)$.

So the equation admits of a Frobenius series of the form $$y=\sum_{n=0}^{\infty}C_n~x^{n+r},\qquad C_0\neq 0 \qquad ..........(2)$$ which converges for all $~x~$.

From $(2)$, $$y'(x)=\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1};\qquad \qquad y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}\qquad .....(3)$$

Substituting $(2)$ and $(3)$ in $(1)$ we get, $$x~\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}+2~\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1}+x~\sum_{n=0}^{\infty}C_n~x^{n+r}=0$$ $$\implies \sum_{n=0}^{\infty}(n+r)~(n+r+1)~C_n~x^{n+r-1}~+~\sum_{n=0}^{\infty}C_n~x^{n+r+1}=0\qquad .....(4)$$

Lowest power of $~x~$ in equation $(4)$ is $~{r-1}~$, so coefficient of $~x^{r-1}~=0$ gives the indicial equation $~r~(r+1)~=0\implies r=0,~-1$

From equation $(4)$ we have the following recursive formula,

$$(n+r+1)~(n+r+2)~C_{n+1}~+~C_{n-1}=0$$ $$\implies C_{n+1}=-\frac{1}{(n+r+1)~(n+r+2)}~C_{n-1}\qquad ........(5)$$

From $(5)$ we have $C_1=C_3=C_5=\cdots =0$

$C_2=-\frac{1}{(r+2)~(r+3)}~C_{0}$

$C_4=-\frac{1}{(r+4)~(r+5)}~C_{2}=\frac{1}{(r+2)~(r+3)~(r+4)~(r+5)}~C_{0}$

$\cdots$

Therefore

$$y(x)=C_0~x^r \left[1-\frac{1}{(r+2)~(r+3)}~x^2+\frac{1}{(r+2)~(r+3)~(r+4)~(r+5)}~x^4-\cdots\right]$$

For $~r=0~$, $$y_1(x)=C_0~ \left[1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots\right]$$ $$\implies y_1(x)=C_0~ \left[1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right]$$

Since $~0-(-1)=1,~$ an integer so the other independent solution of equation $(1)$ is $$y_2(x)=\left[\frac{\partial y}{\partial r}\right]_{r=0}$$ $$\implies y_2(x)=y_1(x)~\log x~+~C_0~\left[\frac{5}{36}~x^2+\cdots\right]$$

General solution is $$y(x)=A~y_1(x)~+~B~y_2(x)\qquad \text{where $~A,~B~$are constants.}$$

Answer 2

You get the equations ($C_{k}=0$ for $k<0$, by construction) $$ (k+r)(k+r+1)C_{k}=-C_{k-2}. $$ The indicial equation for $k=0$ is $r(r+1)=0$. For $r=-1$, the equation for $k=1$ gives also $C_1$ free. All other coefficients are determined by the recursion $$ C_k=-\frac{C_{k-2}}{k(k-1)} $$ so that you get a basis solutions in the even indices $$ y_1(x)=x^{-1}-\frac12x+\frac1{24}x^3-...=\frac{\cos x}{x} $$ and in the odd indices $$ y_2(x)=1-\frac16x^2+\frac1{120}x^4+...=\frac{\sin x}x. $$

The case $r=0$ then just repeats one of these solutions, in that $C_1=0$ and the coefficient recursion $$ C_k=-\frac{C_{k-2}}{k(k+1)} $$ gives multiples of $$ 1-\frac16x^2+...=\frac{\sin x}x $$ that is, of $y_2$.

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Quick check: Setting $u(x)=xy(x)$ gives $u''(x)=xy''(x)+2y'(x)=-xy=-u$, which is indeed the harmonic oscillator equation.