A question about ODE basics.

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• Feb 22nd 2011, 03:57 AM
secondchance
A question about ODE basics.
I am confused about the meaning of the unknown function $y$ in a differential equations.

For example, the Sturm-Liouville equation has the form $\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) + (q(x) + \lambda r(x) ) y = 0$. I notice that the unknown function $y$ is written without its variable $x$, i.e. not written as $y(x)$, whereas the functions $p(x)$, $q(x)$ and $r(x)$ are written with variable $x$.

The following rearrangement of equation is from my lecture notes about orthogonality of eigenfunctions of the Sturm-Liouville equation.
$(\lambda_m - \lambda_n) r(x) \phi_n \phi_m = \phi_m \frac{d}{dx} \left( p(x) \frac{d\phi_n}{dx} \right) - \phi_n \frac{d}{dx} \left( p(x) \frac{d \phi_m}{dx} \right)$
$(\lambda_m - \lambda_n) r(x) \phi_n \phi_m = \frac{d}{dx} \left( p(x) \frac{d \phi_n}{dx} \phi_m - p(x) \frac{d \phi_m}{dx} \phi_n \right)$

Why it is OK to move an unknown function $\phi_m$ of variable $x$ into a derivative operator $\frac{d}{dx}$ by treating it like a constant? I am wondering if I have missed some very basic points about differential equations. Can someone please explain to my why the above rearrangement is possible and when one would write an unknown function as $y$ and when one would write an unknown function as $y(x)$ in an ODE?
• Feb 22nd 2011, 04:57 AM
secondchance
I figured it out. You can't move $\phi_m$ into the derivative operator directly. It turns out that if you differentiate the RHS of the second equation that some terms cancel out and you end up with the RHS of the first equation. I didn't miss anything fundamental and magical.