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

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?

• You already got answers, but I would like to underline a point. The problem here is not the non-constant coefficient. The problem is the singularity at $x=0$. Every well-posedness theorem for second-order ODEs treats equations in the form $y''=f(x, y, y')$. The present equation can be reduced to that form by dividing $x$, but this works only on $(0, \infty)$ or $(-\infty, 0)$, not on $\mathbb R$. Commented Dec 25, 2020 at 16:01

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