# Representing solutions of a second order linear differential equation as the solutions of 2 first order linear differential equations.

Consider $xy''+2y'+xy=0$. Its solutions are $\frac{\cos x}{x},\,\frac{\sin x}{x}$ .

Neither of those solutions (as far as I could find) can be the solutions of a first order linear homogeneous differential equation. However $$\frac{e^{\pm i x}}{x}$$ Can be the solution of a first order linear homogeneous DE ($y'+(x\mp i )y=0$)

Can I always find a first order homogeneous linear DE whose solution also solves a second order homogeneous linear DE?

For example can I find a first order homogeneous linear DE whose solution is a particular linear combination of $J_1$ and $Y_1$ ?

(unrelated: also I'd like to know if there is a way of solving $xy''+2y'+xy=0$ without noticing that it is a spherical bessel function or using laplace transform.)

This may look like cheating, but any $C^1$ function $\psi$ satiafies the first order differential equation $$\psi'+a\,\psi=0\quad\text{with}\quad a=-\frac{\psi'}{\psi}.$$
• Ah, of makes a lot of sense. However I guess what I had in mind was a first order equation that doesn't use fancier functions than the original second order equation. For example for $\sin x /x$ I'd need to have a $\cot x$. Or, $$y'(x)+\frac{y(x) J_1(x)}{J_0(x)}=0$$ has the solution $J_0$ but this isn't really meaningful. Commented Jul 15, 2015 at 21:04
"Can $$\frac{\sin x}{x}$$ be a solution of a linear first order ODE with polynomial coefficients?"
If this is the case, the answer is no. To see why, you can just make a substitution in a equation with generic form $$p(x)y'+q(x)y+r(x)=0$$ to obtain: $$\sin(x)(p-xq)-\cos(x)xp-x^2r=0.$$
By the linear independence of $$\cos(x)$$ and $$\sin(x)$$, we find that $$p=0$$, $$r=0$$ and $$q=0$$.