Why does the ODE $xy''+y'+xy=0$ require only one initial condition?

Question

In order to solve second order differential equations with constant coefficients of the form $$ay''+by'+cy=0$$ we need two initial conditions, for example $y(0)=1$ and $y'(0)=2$.

However, I just crossed the following differential equation with varying coefficients: $$xy''+y'+xy=0 \\y(0)=1$$ known as the Bessel's Equation which can be solved using power series to give the solution $$y(x)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{4^n(n!)^2}x^{2n}.$$ Why did this equation only require one initial condition? is it because it is not with constant coefficients?

Answer 1

Plugging in $x=0$ into the equation gives $y'(0)=0$, so the second initial condition follows from the equation.

Answer 2

This ODE $$xy''+y'+xy=0$$ is invariant under $x\to -x$, so it can have two linearly independent solutions of definite parity. These are $J_0(x)$ and $Y_0(x)$ first one is even but the second one is defined only for $x>0$ and $Y_0(0)=-\infty$ So it cannot have odd prity solution. This aspect is covered by the fact that this ODE has both indicial roots as zrto. Check it by putting $y=x^m$, then equate the coefficient of the lowes power you gwt $m^2=0$. When this happens one solution has logarithmic singularity. You have found $J_0(x)$ as the solution.

The general solution of the ODE can be written as $$y(x)=C_1 J_0(x)+ C_2 Y_0(x)$$ Any two conditions other than $x=0$ can determine $C_1$ and $C_2$ for instance $y(a)=c, y(b)=d.$