on Lagrange's Method. 263 



plain, that it is not easy to account for the mistake; unless 

 indeed his last article help us to a solution of the difficulty. 



He is now puzzled how to deduce the case of equal roots 

 from the usual form of the complete fluent. There is some 

 confusion in what he writes, arising from his not distinguish- 

 ing between the mathematical definition of the complete inte- 

 gral, and a property belonging to it. An algebraic expres- 

 sion which satisfies the differential equation, and contains the 

 requisite number of arbitrary constants, is the complete in- 

 tegral ; and it possesses the property of comprehending every 

 possible case, by varying the arbitrary quantities. It is cer- 

 tain that Mr. Herapath's integral of the third order, when we 

 take in all its three parts, does satisfy the differential equa- 

 tion * ; it also contains three arbitrary constants ; it is there- 

 fore the complete integral : but notwithstanding, he contends 

 that it is not general ; for he makes it fail, particularly in the 

 case of equal roots. Now this proceeds, not from a failure in 

 any method or in any doctrine, but from a failure in Mr. He- 

 rapath, who does not reason correctly in adapting the general 

 expression to the particular case. 



Let us take his own example, p. 212, of the last Number of 

 this Journal, viz. 



= ce rx -c t e r * x +e rx /Xe - rx dx-e r * x /Xe- r * X dx 

 " r — r, 



When the two roots are real and unequal, or when they are 

 imaginary, there is no difficulty; but the puzzle is, what must 

 be done when the roots are equal, for then the integral is in- 

 finite. For the sake of simplicity, let r x — r = co, then r, m 

 r -f cy; also put c t = c + hoo: then, by substitution, 



e rx + » x (c + h u +SXe- rx -"' x d X )-e rx (c+fXe- rx dx) 

 y - _ 



This is only another form of the complete integral, the arbi- 

 trary constants being c and h; but it has the advantage of 

 bringing out the correct value of y when the roots are equal, 



or co = 0. If we substitute the expansions of e ax and e~ ax 9 

 we shall obtain, supposing co = 0, 



y = e rx (ji -f- ex + xfe~~ r *X.dx —fe~ "Xxd*); 



b\xt^ xfe" *Kdx — fe Xxdx =fdxfe~ Hdx; 



wherefore, y = e r x yh + c x 4- fd xfe ~~ r x X d xj 9 



* See the calculation in this Journal for February last, p. 97. 



which 



