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Consider the differential equation:-

$a \phi + (bD^3 - cD)w =0$, where $a, b$ and $c$ are constants, $D$ denotes the differential operator $\dfrac{d}{dx}$, and $w$ is a function of $x$.

I'm defining $w = Lw'$ and $x=Lx'$, where $L$ is a constant.

I'm trying to obtain $\phi$ in terms of $x'$. But I've two questions that pop into my mind immediately:-

$1.$ How do I change the differential operator from $\dfrac{d}{dx}$ to $\dfrac{d}{dx'}$, so that I can obtain $\phi$ correctly in terms of $x'$?

$2.$ Let's say I'm keeping the differential operator as such, and differentiating $w$ with respect to $x$. After the differentiation, if I substitute $x$ with $Lx'$, is $\phi$ the same as the one obtained by changing the differential operator?

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Here is a start. First make the change of the dependent variable $w=Lz$ (I used z instead of w' to avoid confusion with derivative)

$$ w=Lz \implies D^n w= L D^n z,\quad D=\frac{d}{dx}, $$

so, the differential equation becomes

$$ a \phi(x) + L(bD^3 - cD)z =0 \longrightarrow (1).$$

Now, we use the other change of variables $x=Lt$ (again I let $t=x'$ avoiding the confusion) in $(1)$ as

$$ \frac{dz}{dx} = \frac{dz}{dt}\frac{dt}{dx} = \frac{1}{L} \frac{dz}{dt} $$

$$ \implies D^2 z = \frac{d^2z}{dx^2} = \frac{d}{dx}\left(\frac{1}{L}\frac{dz}{dt}\right)\frac{dt}{dx} = \frac{d}{dt}\left(\frac{1}{L}\frac{dz}{dt}\right)\frac{dt}{dx}=\frac{1}{L^2}\frac{d^2z}{dt^2}$$

$$ \implies D^3 z = \frac{1}{L^3}\frac{d^3z}{dt^3}. $$

Now, just go back and make substitutions in $(1)$. I think you can do that.

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    $\begingroup$ Thank you so much! I understand it now. $\endgroup$ – Train Heartnet Jan 27 '14 at 14:23
  • $\begingroup$ @downvoter: Is there a reason for the downvote? $\endgroup$ – Mhenni Benghorbal Feb 12 '14 at 18:59

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