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I am very stuck on a homework problem involving series solutions and 2nd order ODE. Could anyone point me towards a solution?

Consider the ODE $$ xy'' + y' - y = 0 $$ 0 is a singular point for the differential equation, but there is a solution that is analytic at 0. Find the series representation centered at 0 for this solution.

Any help at all would be greatly appreciated.

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  • $\begingroup$ Try Frobenius method. $\endgroup$ Nov 9, 2016 at 5:01
  • $\begingroup$ I've tried that and just get exponents at the singularity of r=0, which in turn results in a power series of nothing but zeroes. Am I doing something wrong here? $\endgroup$
    – dmilani
    Nov 9, 2016 at 6:12
  • $\begingroup$ Same question with partial answer on the "Howto": math.stackexchange.com/q/1579870/115115 $\endgroup$ Jul 15, 2019 at 12:09

1 Answer 1

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

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

Lowest power of $~x~$ in equation $(4)$ is $~{r-1}~$, so coefficient of $~x^{r-1}~=0$ gives the indicial equation $~r^2~=0\implies r=0,~0$

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

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

From $(5)$ we have

$C_1=\frac{1}{(r+1)^2}~C_{0}$

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

$C_3=\frac{1}{(r+3)^2}~C_{2}=\frac{1}{(r+1)^2~(r+2)^2~(r+3)^2}~C_{0}$

$\cdots$

Therefore

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

For $~r=0~$, $$y_1(x)= \left[1+~x~+\frac{x^2}{4}+\frac{x^3}{36}+\cdots\right]$$ $$\implies y_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{(n!)^2}=J_0 (2\sqrt{x})$$ $J_0(X)~$ is the modified Bessel function of first kind and order $~0~$.

The other independent solution of equation $(1)$ is $$y_2(x)=\left[\frac{\partial y}{\partial r}\right]_{r=0}$$ $$\implies y_2(x)=y_1(x)~\log x~-~\left[2~x+\frac{3}{4}~x^2~+\cdots\right]$$ $$\implies y_2(x)=Y_0 (2\sqrt{x})$$ $Y_0(X)~$ is the modified Bessel function of second kind and order $~0~$.

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

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  • $\begingroup$ Use \tag1, \tag{12a}, \tag{eqn $1$} etc. for uniformly placed equation numbers. $\endgroup$ Jul 15, 2019 at 11:43
  • $\begingroup$ Thanks for your valuable suggestion @LutzL $\endgroup$
    – nmasanta
    Jul 15, 2019 at 11:44

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