Mathematical Methods for Physics and Engineering : A Comprehensive Guide

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS

Solve

6 y^2 p^2 +3xp−y=0. (14.32)

This equation can be solved forxexplicitly to give 3x=(y/p)− 6 y^2 p. Differentiating both
sides with respect toy, we find

3

dx

dy

=

3

p

=

1

p

−

y

p^2

dp

dy

− 6 y^2

dp

dy

− 12 yp,

which factorises to give

(

1+6yp^2

)(

2 p+y

dp

dy

)

=0. (14.33)

Setting the factor containingdp/dyequal to zero gives a first-degree first-order equation
inp, which may be solved to givepy^2 =c. Substituting forpin (14.32) then yields the
general solution of (14.32):

y^3 =3cx+6c^2. (14.34)

If we now consider the first factor in (14.33), we find 6p^2 y=−1 as a possible solution.
Substituting forpin (14.32) we find the singular solution

8 y^3 +3x^2 =0.

Note that the singular solution contains no arbitrary constants and cannot be found from
the general solution (14.34) by any choice of the constantc.

Solution method.Write the equation in the form (14.31) and differentiate both

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

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

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

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

fashion will instead lead to singular solutions.

14.3.3 Equations soluble fory

Equations that can be solved fory, i.e. are such that they may be written in the

form

y=F(x, p), (14.35)

can be reduced to first-degree first-order equations inpby differentiating both

sides with respect tox,sothat

dy

dx

=p=

∂F

∂x

+

∂F

∂p

dp

dx

.

This results in an equation of the formG(x, p) = 0, which can be used together

with (14.35) to eliminatepand give the general solution. An additional (singular)

solution to the equation is also often found.