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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.)

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This may look like cheating, but any $C^1$ function $\psi$ satiafies the first order differential equation $$ \psi'+a\,\psi=0\quad\text{with}\quad a=-\frac{\psi'}{\psi}. $$

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 15, 2015 at 21:04
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Maybe when you say "doesn't use fancier functions" you want to say "polynomial coeficients". That is:

"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 $q=0$.

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