Page:A History of Mathematics (1893).djvu/226

 "Now, by supposition, $\scriptstyle{x^3-ax^2+axy-y^3=0}$, which therefore, being expunged and the remaining terms being divided by 0, there will remain

But whereas zero is supposed to be infinitely little, that it may represent the moments of quantities, the terms that are multiplied by it will be nothing in respect of the rest (termini in eam ducti pro nihilo possunt haberi cum aliis collati); therefore I reject them, and there remains

as above in Example I." Newton here uses infinitesimals.

Much greater than in the first problem were the difficulties encountered in the solution of the second problem, involving, as it does, inverse operations which have been taxing the skill of the best analysts since his time. Newton gives first a special solution to the second problem in which he resorts to a rule for which he has given no proof.

In the general solution of his second problem, Newton assumed homogeneity with respect to the fluxions and then considered three cases: (1) when the equation contains two fluxions of quantities and but one of the fluents; (2) when the equation involves both the fluents as well as both the fluxions; (3) when the equation contains the fluents and the fluxions of three or more quantities. The first case is the easiest since it requires simply the integration of $\scriptstyle{\frac{dy}{dx}=f(x)}$, to which his "special solution" is applicable. The second case demanded nothing less than the general solution of a differential equation of the first order. Those who know what efforts were afterwards needed for the complete exploration of this field in analysis, will not depreciate Newton's work