Mathematical Methods for Physics and Engineering : A Comprehensive Guide

14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONS

Solve

xp^2 +2xp−y=0. (14.36)

This equation can be solved foryexplicitly to givey=xp^2 +2xp. Differentiating both
sides with respect tox, we find

dy

dx

=p=2xp

dp

dx

+p^2 +2x

dp

dx

+2p,

which after factorising gives

(p+1)

(

p+2x

dp

dx

)

=0. (14.37)

To obtain the general solution of (14.36), we consider the factor containingdp/dx.This
first-degree first-order equation inphas the solutionxp^2 =c(see subsection 14.3.1), which
we then use to eliminatepfrom (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 simple
solutionp=−1. Substituting this into (14.36) then gives

x+y=0,

which is a singular solution to (14.36).

Solution method.Write the equation in the form (14.35) and differentiate both

sides with respect tox. Rearrange the resulting equation into the formG(x, p)=0,

which can be used together with the original ODE to eliminatepand so give the

general solution. IfG(x, p)can be factorised then the factor containingdp/dxshould

be used to eliminatepand give the general solution. Using the other factors in this

fashion will instead lead to singular solutions.

14.3.4 Clairaut’s equation

Finally, we consider Clairaut’s equation, which has the form

y=px+F(p) (14.39)

and is therefore a special case of equations soluble fory, as in (14.35). It may be

solved by a similar method to that given in subsection 14.3.3, but for Clairaut’s

equation the form of the general solution is particularly simple. Differentiating

(14.39) with respect tox, we find

dy

dx

=p=p+x

dp

dx

+

dF

dp

dp

dx

⇒

dp

dx

(

dF

dp

+x

)

=0. (14.40)

Considering first the factor containingdp/dx, we find

dp

dx

=

d^2 y

dx^2

=0 ⇒ y=c 1 x+c 2. (14.41)