## Folio 174: AAL to ADM

**[174r]** [(mostly) in AAL's hand]

Theorem. Page 199.

If \(N\) be a function of \(x\) and \(y\), giving \(\frac{dN}{dx}=p+q\frac{dy}{dx}\)

then the equation \(\frac{du}{dx.dy}=V.\frac{dN}{dx.dy}\) is incongruous &

self-contradictory, except upon the assumption

that \(u\) is, as to \(x\) and \(y\), a function of \(N\);

or contains \(x\) and \(y\) only thro' \(N\) .

Let \(N=\psi(x,y)\) give \(y=\chi(N,x)\), and

suppose, if possible, that the substitution of

this value of \(y\) in \(u\) gives \(u=\beta(N,x)\), \(x\)

not disappearing with \(y\) . Then \(x\) and \(y\)

varying

\(\frac{du}{dx.dy}=\frac{d\beta}{dN}.\frac{dN}{dx}+\frac{d\beta}{dN}.\frac{dN}{dy}+\frac{d\beta}{dx}\) [in above line, \(\frac{du}{dx.dy}\) is crossed through in pencil, and '\( 1\) ' written above; </br>'\( =\frac{du}{dx}+\frac{du}{dy}\) ' added in pencil at end of line --- in ADM's hand?]

\(=\frac{d\beta}{dN}\cdot\left(\frac{dN}{dx}+\frac{dN}{dy}\right)+\frac{d\beta}{dx}=\frac{d\beta}{dN}\cdot\frac{dN}{dx.dy}+\frac{d\beta}{dx}=\)

\(=V.\frac{dN}{dx.dy}\) , which equation being

universal\, is true on the supposition that \(x\)

does not vary\, or that \(\frac{d\beta}{dx}=0\) . This gives \(\frac{d\beta}{dN}=V\);

or \(\frac{du}{dx.dy}=V\frac{dN}{dx.dy}+\frac{d\beta}{dx}=V\frac{dN}{dx.dy}\)

because \(\frac{d\beta}{dN}\) and \(V\) being independent of the variations

&c, &c. Hence \(\frac{d\beta}{dx}=0\) always; or \(\beta\) does not

contain \(x\) directly, &c.

I think the above is correct. I cannot see**[174v] ** the use (page 200) of introducing \(t\) in

the proof __there__ given . Is it possible that

I have committed an error in my original

understanding of the __ennunciation__ [*sic*] of the Theorem;

& that the \(du\) ['of the equation' crossed out] and the \(dN\)

of the equation \(du=V.dN\) do not mean

the \(du\) and \(dN\) derived from differentiating__with respect to the quantities \(x\) and \(y\) __,

a__lready__ introduced ; but with respect

to ['some' crossed out] __other__ given quantity?

I suspect so .

[the following appears underneath in pencil --- still in Ada's hand]}

\(u=\beta(N, x)\)

\(\frac{du}{dx}=\frac{d\beta}{dN}\frac{dN}{dx}+\frac{d\beta}{dx}\)

\(\frac{d^nu}{dx.dy}=\)

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