$xy''-y=0$ with initial conditions $y(0)=0$ and $y'(0)=1$.

Question

$xy''-y=0$ with $y(0)=0$ and $y'(0)=1$

Hello! I'm currently working on this problem in my differentials course and I've come up on an issue..

My solution utilizes the Frobenius method and I found the two roots of the indicial equation to be $1$ and $0$.

Given this I solved for my two solutions but one of my linearly independent solutions looks like this $$y(x)=ay_1(x)\log(x) + \sum b_nx^n$$ I'm not sure how to apply these initial conditions as $\log(0)$ is undefined, any help would be appreciated

$$y_1(x)=x\sum a_n(-1)^n\frac{x^n}{(n+1)!\,n!}$$

Answer 1

Not an answer to the question, just for information. Too long to be edited in the comment section.

$$xy''-y=0$$ is an ODE of the Bessel kind.

It is easy to find the general solution in terms of Bessel functions : $$y(x)=c_1\sqrt{x}I_1(2\sqrt{x})+c_2\sqrt{x}K_1(2\sqrt{x}) \quad\text{if}\quad x>0$$ $$y(x)=c_1\sqrt{-x}J_1(2\sqrt{-x})+c_2\sqrt{-x}Y_1(2\sqrt{-x}) \quad\text{if}\quad x<0$$ $J_1,Y_1$ are the first order Bessel function of first and second kind. $I_1,K_1$ are the first order modified Bessel function of first and second kind.

Case $x>0$:

The series expansion for $x\to 0^+$ is : $$y(x)=c_1\left(x+\frac{x^2}{2}+O(x^3)\right)+c_2\left(\frac12+\frac12(\ln(x)+2\gamma-1)x+O\left(x^2\ln(x)\right) \right)$$ Thus, the condition $y(0)=0$ implies $c_2=0$ and then : $$y(x)=c_1\left(x+O(x^2)\right)$$ $$y'(x)=c_1\left(1+O(x)\right)$$ The condition $y'(0)=1$ implies $c_1=1.$ So, the solution is : $$y(x)=\sqrt{x}I_1(2\sqrt{x})\qquad x>0$$ Similarly, one obtains : $$y(x)=\sqrt{-x}J_1(2\sqrt{-x})\qquad x<0$$

This functions expressed on the form of series agree with the result already given in the preceding answers.

Answer 2

That formula should be $$y_1(x)=x\cdot a_0\sum_{n=0}^\infty\frac{(-1)^nx^n}{(n+1)!n!},$$ as you replace all $a_n$ in $y_1(x)=\sum a_nx^n$ with multiples of $a_0$.

Accidentally the first initial condition is automatically satisfied ($y(0)=1$ would be impossible for any solution) and with $a_0=1$ you also get the second one.