By V. I. Smirnov and A. J. Lohwater (Auth.)

**Read Online or Download A Course of Higher Mathematics. Volume II PDF**

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**Extra info for A Course of Higher Mathematics. Volume II**

**Sample text**

88) We start by finding the orthogonal trajectory. e. the slopes of the tangents to the trajectory are the reciprocals, with reversed sign, of the slopes of the tangents to the given family. Hence it follows that, to obtain the differential equation of the orthogonal trajectory, we must replace y'by (— \\y') in the differential equation of the given family. Finding the orthogonal trajectory thus reduces t o integration of the equation: *(*•*>-£) = °. where yx is the required function of x. We now turn t o the general problem of isogonal trajectories.

This substitution is of great importance in graphical statics. Examples. 1. A beam is constrained at the end O and subjected to a con centrated vertical force P at the end L (Fig. 20); the weight of the beam can be neglected. e. o = ° and y' lx-o = o, so that we find [15]: y-1 <*-«-£-»-{> __
__

__We get two cases on equating each factor to zero. The case dp = 0 gives p = C, where (7 is an arbitrary constant; substitution of p = 0 in equation (67) again gives us the general solution (66). In the second case we have the equation: x + __