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Dividing the first of these by the second we find

Dividing the original equation by the first we have

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By means of these values, eliminating p and q, we find the singular solution to be

arber na y" ze = (-)"+b A."

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If the partial differential equation be of the second order and we put


dz dal dxdy dys the conditions which must be satisfied in order that an equation should be the singular solution of the first order of the equation U = 0 are

dU dU

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If the function is to be a singular solution belonging to the final integral, it must in addition satisfy the equations

du du

= 0.
(24) Let the given equation be

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dU Here

= 0 is the only condition, and we have

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Comparing this with the given equation, we find

p- ^, = 0, from which r=1ta (-11 2) = 0, which satisfies both U = 0 and = 0: therefore

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p-11-0 is the singular solution required. The integral of this is

% = (1 + x) °(). Poisson, Jour. de l'Ecole Polyt. Cah. xIII. p. 113. (25) Let the equation be

que - € + (-3) (- ~ – y) = 0. It will be found that

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it is therefore a singular solution corresponding to the final integral.




Sect. 1. Quadrature of Plane Areas. When an area is referred to rectangular co-ordinates x and y, the double integral Sfdxdy taken between the proper limits gives the value of the area. One of the integrations may always be performed, so that we have either

Sydx + C or fædy + C, and these integrals are to be taken between the limits of y or X, which form the boundaries of the area. If we take the first of these expressions, the limiting values of y must either be constants or functions of x given by the equation to the bounding curve: therefore on substituting these values we obtain a function of x alone, which is to be integrated, and taken between the limits of that variable which are required by the problem. If after the first integration we suppose C = 0, the integral A = sydx expresses the area included between the axis of x, the curve, and two ordinates corresponding to the limits of x.

In taking the integral sy dx between the final limits of X, it is necessary that the interval should not contain a value of x which causes y to vanish or become infinite, as in that case we might be led to an erroneous conclusion. Thus if we suppose a curve to be symmetrically situate in the first and third quadrants, and to intersect the axis at the origin; and if we were to integrate from x = a to x = – a we should obtain zero as our result, instead of finding the area to be double of that from 2 = 0 to x = a, or that from x = 0 to Q = - a. Therefore when any interval from a to b contains a value c of x which makes y vanish or become infinite, we must break it up into two intervals, one from b to c and the other from c to a, and add the integrals corresponding to these. In like manner if the interval contain several values of a which make y vanish or become infinite, we must split it up into as many smaller intervals, each having one of these values of x as a limit, and add them all together.

If the co-ordinates be not rectangular, and a be the angle between them, we must multiply the integral by sin a to obtain the value of the area.

Ex. (1) If we take the general equation to a parabola of any order

Yh+ = a" 2",

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Determining the constant by the condition that the area vanishes when x = a, we have

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