given $ax^2y''+bxy'+cy=0$ a,b, and c are real. x is positive. I want to show that  $$a \frac{d^2y}{dv^2} +(b-a)\frac{dy}{dv} +cy =0.$$ This is a problem in a elementary text, I have found a more rigours method online, but I'm trying to make sense of the last step in my own less rigours idea. 

Using the coefficients $a$, $b$, and $c$, I observe that if this is true then $$xy'=\frac{dy}{dv}\quad\text{and}\quad x^2y''= \frac{d^2y}{dv^2} - \frac{dy}{dv},$$ so if I show the two statements above are true I will have justified the subsitution.

Let $\ln x = v$ $$\frac{1}{x} = \frac{dv}{dx}$$ $\ln$ is one to one so it has an inverse and  $$\begin{align*} x &= \frac{dx}{dv}\\ xy'&= \frac{dx}{dv} \frac{dy}{dx}\&&\text{by the chain rule}\\ xy'&= \frac{dy}{dv}. \end{align*}$$ Great one down one to go. Continuing with the above... $$x \frac{dy}{dx} = \frac{dy}{dv}.$$ I take the derivative with respect to $x$ of both sides, using the product rule on the right. $$x \frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{d}{dx}  \left( \frac{dy}{dv} \right)$$ I dont feel so good about the right side, but I keep going anyway.

$$x \frac{d^2y}{dx^2} = \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dx}$$ since $x$ is $e^v$, $x$ is its own derivative with respect to $v$. I multiply through by $x$: $$x^2 \frac{d^2y}{dx^2} = x \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dx} \frac{dv}{dx}$$ chain rule on the far right.

$$x^2 \frac{d^2y}{dx^2} = x \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dv}$$ I'm so close but I'm stuck! all I want to say is: $$x^2y''= \frac{d^2y}{dv^2} - \frac{dy}{dv}$$ and it is very sugestive to write: $$x^2 \frac{d^2y}{dx^2} = \frac{dx}{dv} \frac{d}{dx}  \left( \frac{dy}{dv} \right) - \frac{dy}{dv}$$ but what could it mean to cross out the $dx/dx$? that is in my way? as I was told before here such operations are "dubious" though I'm still trying to grasp why.

But I know this is true so it must be the case that for these functions $$\frac{dx}{dv} \frac{d}{dx}  \left( \frac{dy}{dv} \right) = \frac{d^2y}{dv^2}$$ perhaps I can say that the kind of functions that work in the above are just the kind I'm working with and then I would be done? 

  • $\begingroup$ Is the original equation $ax^2y''+bxy'+cy=0$? I think you dropped an $x$ in the second term. $\endgroup$ Jun 23 '11 at 20:22
  • $\begingroup$ What happened to my shorter version? $\endgroup$ Jun 23 '11 at 20:24
  • $\begingroup$ Yes, I dropped an x. $\endgroup$ Jun 23 '11 at 20:25
  • 1
    $\begingroup$ The last step is just the chain rule again. $\endgroup$ Jun 23 '11 at 20:25
  • 2
    $\begingroup$ @Not: I believe what Raskolnikov is saying: If $f = \frac{dy}{dv}$ then $\frac{df}{dx} = \frac{df}{dv} \times \frac{dv}{dx}$. So with $f = \frac{dy}{dv}$ we get $\frac{d^2y}{dv^2} \times \frac{dv}{dx}$. $\endgroup$
    – Aryabhata
    Jun 23 '11 at 20:50

Start with $$a{x^2}y'' + bxy' + cy = 0$$

which is an Euler-Cauchy DE of second degree. Now make the change of variables,

$$x = e^z$$

From here you get that

$$dx = e^z dz$$

So now you have

$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}}\frac{{dz}}{{dx}}=\frac{{dy}}{{dz}}\frac{1}{x}$$


$$bx\frac{{dy}}{{dx}} = b\frac{{dy}}{{dz}}$$

Which is what you correctly derived.

For the second equality go this way

$$\eqalign{ & x = {e^z} \cr & \frac{{dy}}{{dx}} = {e^{ - z}}\frac{{dy}}{{dz}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {{e^{ - z}}\frac{{dy}}{{dz}}} \right) \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dz}}\left( {{e^{ - z}}\frac{{dy}}{{dz}}} \right)\frac{{dz}}{{dx}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \left( {{e^{ - z}}\frac{{{d^2}y}}{{d{z^2}}} - {e^{ - z}}\frac{{dy}}{{dz}}} \right)\frac{{dz}}{{dx}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2z}}\left( {\frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}}} \right) \cr & {x^2}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}} \cr} $$

Notice that what is used is the chain rule, in a rather "algebraic" way.

You can prove by induction that

$${x^n}{D^n}y = \mathcal{D}\left( {\mathcal{D} - 1} \right) \cdots \left( {\mathcal{D} - n + 1} \right)y$$

Where $\mathcal{D}=\dfrac{d}{dz}$ and $D = \dfrac{d}{dx}$


With respect to $\frac{dx}{dx}$, the reason such operations are "dubious" is because there is a subtle difference between differential division and applying the differential operator ($\frac{d}{dx}$), so even though they are written the same way, you could argue that it should be written $\frac{dx}{dx}$ and $\frac{d}{dx}x$. The distinction is clear in higher order differentials: $\frac{dy^2}{dx^2}$ vs $\frac{d^2y}{dx^2}$($=\left(\frac{d}{dx}\right)^2y=\frac{d^2}{dx^2}y$).


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