Mathematical Methods for Physics and Engineering : A Comprehensive Guide

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS

14.2.2 Exact equations

Anexactfirst-degree first-order ODE is one of the form

A(x, y)dx+B(x, y)dy= 0 and for which

∂A

∂y

=

∂B

∂x

. (14.4)

In this caseA(x, y)dx+B(x, y)dy is an exact differential,dU(x, y)say(see

section 5.3). In other words

Adx+Bdy=dU=

∂U

∂x

dx+

∂U

∂y

dy,

from which we obtain

A(x, y)=

∂U

∂x

, (14.5)

B(x, y)=

∂U

∂y

. (14.6)

Since∂^2 U/∂x∂y=∂^2 U/∂y∂xwe therefore require

∂A

∂y

=

∂B

∂x

. (14.7)

If (14.7) holds then (14.4) can be writtendU(x, y) = 0, which has the solution

U(x, y)=c,wherecis a constant and from (14.5)U(x, y) is given by

U(x, y)=

∫

A(x, y)dx+F(y). (14.8)

The functionF(y) can be found from (14.6) by differentiating (14.8) with respect

toyand equating toB(x, y).

Solve

x

dy

dx

+3x+y=0.

Rearranging into the form (14.4) we have

(3x+y)dx+xdy=0,

i.e.A(x, y)=3x+yandB(x, y)=x.Since∂A/∂y=1=∂B/∂x, the equation is exact, and
by (14.8) the solution is given by

U(x, y)=

∫

(3x+y)dx+F(y)=c 1 ⇒

3 x^2

2

+yx+F(y)=c 1.

DifferentiatingU(x, y) with respect toyand equating it toB(x, y)=xwe obtaindF/dy=0,
which integrates immediately to giveF(y)=c 2. Therefore, lettingc=c 1 −c 2 ,thesolution
to the original ODE is

3 x^2

2

+xy=c.