Mathematical Methods for Physics and Engineering - Matematica.NET

14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONS◮Solvexp 2 +2xp − y =0. (14.36)This equation can be solved for y explicitly to give y = xp 2 +2xp. Differentiating bothsides with respect to x, we finddydxwhich after factorising gives= p =2xpdpdx + p2 +2x dpdx +2p,((p +1) p +2x dp )=0. (14.37)dxTo obtain the general solution of (14.36), we consider the factor containing dp/dx. Thisfirst-degree first-order equation in p has the solution xp 2 = c (see subsection 14.3.1), whichwe then use to eliminate p from (14.36). Thus we find that the general solution to (14.36)is(y − c) 2 =4cx. (14.38)If instead, we set the other factor in (14.37) equal to zero, we obtain the very simplesolution p = −1. Substituting this into (14.36) then givesx + y =0,which is a singular solution to (14.36). ◭Solution method. Write the equation in the form (14.35) and differentiate bothsides with respect to x. Rearrange the resulting equation into the form G(x, p) =0,which can be used together with the original ODE to eliminate p and so give thegeneral solution. If G(x, p) can be factorised then the factor containing dp/dx shouldbe used to eliminate p and give the general solution. Using the other factors in thisfashion will instead lead to singular solutions.14.3.4 Clairaut’s equationFinally, we consider Clairaut’s equation, which has the formy = px + F(p) (14.39)and is therefore a special case of equations soluble for y, as in (14.35). It may besolved by a similar method to that given in subsection 14.3.3, but for Clairaut’sequation the form of the general solution is particularly simple. Differentiating(14.39) with respect to x, we finddydx = p = p + x dpdx + dF dpdp dx⇒dp ( ) dFdx dp + x =0. (14.40)Considering first the factor containing dp/dx, we finddpdx = d2 ydx 2 =0 ⇒ y = c 1x + c 2 . (14.41)483