Mathematical Methods for Physics and Engineering - Matematica.NET

15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONSbut also writed 2 ydx 2 = dpdx = dy dpdx dy = p dpdyd 3 ydx 3 = d (p dp )= dy ddx dy dx dy(p dp )( ) 2= p 2 d2 p dpdy dy 2 + p , (15.78)dyand so on for higher-order derivatives. This leads to an equation of one orderlower.◮Solve( ) 21+y d2 y dydx + =0. (15.79)2 dxMaking the substitutions dy/dx = p and d 2 y/dx 2 = p(dp/dy) we obtain the first-orderODE1+yp dpdy + p2 =0,which is separable and may be solved as in subsection 14.2.1 to obtain(1 + p 2 )y 2 = c 1 .Using p = dy/dx we therefore have√p = dydx = ± c 2 1 − y2,y 2which may be integrated to give the general solution of (15.79); after squaring this reads(x + c 2 ) 2 + y 2 = c 2 1. ◭Solution method. If the ODE does not contain x explicitly then substitute p =dy/dx, along with the relations for higher derivatives given in (15.78), to obtain anequation of one order lower, which may prove easier to solve.15.3.3 Non-linear exact equationsAs discussed in subsection 15.2.2, an exact ODE is one that can be obtained bystraightforward differentiation of an equation of one order lower. Moreover, thenotion of exact equations is useful for both linear and non-linear equations, sincean exact equation can be immediately integrated. It is possible, of course, thatthe resulting equation may itself be exact, so that the process can be repeated.In the non-linear case, however, there is no simple relation (such as (15.43) forthe linear case) by which an equation can be shown to be exact. Nevertheless, ageneral procedure does exist and is illustrated in the following example.519