By V. I. Smirnov and A. J. Lohwater (Auth.)
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19 the beam either to the left or to the right of N, and calculate the bending moment M(x), equal to the sum of the moments about the neutral line of section N of all the external forces acting on the neglected part, these moments being reckoned positive if, in the case of neglecting the left-hand part, they have a counter clockwise rotation, or in the case of neglecting the right-hand part, they have 50 ORDINARY DIFFERENTIAL EQUATIONS [16 a clockwise rotation. The differential equation of the bent axis of the beam now reads: El -g-^Affc), (24) where E is the modulus of elasticity, and I the moment of inertia of the crosssection considered about the neutral line contained in it.
The envelope of a family of curves, and singular solutions. We have already had two examples in which singular solutions were obtained in addition to the general solution. The general solution in the example of  consisted of the family of circles (a? — C)2 + y2 = a2 (77) with centres on OX and of fixed radius a. The two lines y = ± a , parallel to OX, were singular solutions. Any given point of these lines is a point of contact with a circle of family (77) (Fig. 10). e. the family of straight lines was a family of tangents to the astroid.
E. the direction of each of lines (66) is the same as the constant tangential direction defined by the differential equation at points of the line. e. the family of isoclines (66) is at the same time the family of the general solution of (65). We now indicate a second method of obtaining the general solution of equation (65), whereby the singular solution of the equation is 8] 29 CLAIRATJT'S EQUATION found, as well as its general solution. We use the notation y' = p, and re-write (65): y = xp +
A Course of Higher Mathematics. Volume II by V. I. Smirnov and A. J. Lohwater (Auth.)