g2esebyy7

2022-04-29

I have the differential equation

$\frac{d}{dx}\left(p\left(x\right)\frac{df}{dx}\right)+\cdots =0$

and I want to perform a generic change of variable from x to$y=y\left(x\right)$ .

and I want to perform a generic change of variable from x to

Kendrick Fritz

Beginner2022-04-30Added 12 answers

Step 1

It may be illuminating to work using functional notation rather than Leibniz notation. The expression you are concerned with rewriting is

$(p\xb7f\text{'})\text{'}=p\text{'}\xb7f\text{'}+p\xb7f\text{'}\text{'}.$

Let $p=P\circ y$ and $f=F\circ y.$. Thus $p\text{'}=(P\text{'}\circ y)\xb7y\text{'},$ and $f\text{'}=(F\text{'}\circ y)\xb7y\text{'},$ and

$f\text{'}\text{'}=(F\text{'}\circ y)\text{'}\xb7y\text{'}+(F\text{'}\circ y)\xb7y\text{'}\text{'}=(F\text{'}\text{'}\circ y)\xb7(y\text{'}{)}^{2}+(F\text{'}\circ y)\xb7y\text{'}\text{'}.$

As such,

$(p\xb7f\text{'})\text{'}=(P\text{'}\circ y)\xb7y\text{'}\xb7(F\text{'}\circ y)\xb7y\text{'}+(P\circ y)\xb7(F\text{'}\text{'}\circ y)\xb7(y\text{'}{)}^{2}+(P\circ y)\xb7(F\text{'}\circ y)\xb7y\text{'}\text{'}.$

obettyQuokeperg6

Beginner2022-05-01Added 14 answers

Step 1

It is correct that

$\frac{\mathrm{d}}{\mathrm{d}x}[p\xb7\frac{\mathrm{d}f}{\mathrm{d}x}]=\frac{\mathrm{d}}{\mathrm{d}y}[p\xb7\frac{\mathrm{d}F}{\mathrm{d}y}\xb7\frac{\mathrm{d}y}{\mathrm{d}x}]\xb7\frac{\mathrm{d}y}{\mathrm{d}x}.$

Here, you must use the product rule:

$\frac{\mathrm{d}}{\mathrm{d}y}[p\xb7\frac{\mathrm{d}F}{\mathrm{d}y}\xb7\frac{\mathrm{d}y}{\mathrm{d}x}]\xb7\frac{\mathrm{d}y}{\mathrm{d}x}$

$=\frac{\mathrm{d}}{\mathrm{d}y}[p\xb7\frac{\mathrm{d}F}{\mathrm{d}y}](\frac{\mathrm{d}y}{\mathrm{d}x}{)}^{2}+p\xb7\frac{\mathrm{d}F}{\mathrm{d}y}\xb7\frac{\mathrm{d}y}{\mathrm{d}x}\xb7\frac{\mathrm{d}}{\mathrm{d}y}(\frac{\mathrm{d}y}{\mathrm{d}x})$

The secret here is that

$\frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{d}y}}\left(\frac{{\displaystyle \mathrm{d}y}}{{\displaystyle \mathrm{d}x}}\right)$

$=\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}\right)$

$=-\frac{\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{y}^{2}}}{(\frac{\mathrm{d}y}{\mathrm{d}x}{)}^{2}}.$

The problem here is that, in actuality, you really cannot do such arbitrary changes of variable without accounting for y' in the change of variables.

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