30 Prof. Challis on a new Integration of Differential 



A for the quantity in brackets, with equal reason 



A = 0, and Ap=0. 

 If it might be assumed that these are identical equations, it 

 woultl suffice to integrate one of thern. But, in fact, although 

 in some instances both are immediately integrable and give 

 identical relations between x and y, in others only one, or 

 neither of the two, is immediately integrable; and it cannot, 

 therefore, be antecedently affirmed that they are identical equa- 

 tions. Hence it is necessary to take account of results dedu- 

 cible from them either separately or conjointly. I propose, first, 

 to adduce an instance in which the integrals of the two equa- 

 tions are identical. 



Problem I. Required to connect two fixed points by a curve 

 of given length so that the area bounded by the curve, the or- 

 dinates of the fixed points, and the axis of abscissae shall be a 

 maximum. 



According to Lemma I. the area thus defined is equal to 

 \ydx taken from the abscissa of one extremity of the curve to the 

 abscissa of the other ; and by Lemma II. the length of the curve 



is \\/l+p 2 dx between the same limits, p being put for — . 

 J dx 

 Hence by the Calculus of Variations, a being an arbitrary con- 

 stant, 



8f(y + a^l+p*)dx=Q. 



Putting V for the quantity in brackets, we have 



AT dV , , -o dV ap 



N = -j— == lj and P = -~ = 



Hence 



dy dp Vl+/ 



»-£) 



d¥ \ _ -\ _ d ap 



dx vi+y 

 p 

 +P' 



ap 

 Consequently Adx== dx — d. — —- — 2 = 0; and by integrating, 



ap 



x + c = 



Vl-ff 2 

 Hence, by integrating again, 



(x + cy+(y + c')* = a*. 

 Also 



A P dx= P J X -p . d . ^JL^ = dy- ^E&_ =0; 



