# Series Solution to a 2nd order ODE

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.

• Try Frobenius method. Nov 9, 2016 at 5:01
• 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? Nov 9, 2016 at 6:12
• Same question with partial answer on the "Howto": math.stackexchange.com/q/1579870/115115 Jul 15, 2019 at 12:09

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

$$~x=0~$$ 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 $$y(x)=A~y_1(x)~+~B~y_2(x)\qquad \text{where ~A,~B~are constants.}$$

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