Kenneth Kuttler

29.6. THE INTEGRATING FACTOR 561

Then integrating the first term with respect to x, the scalar potential is of the form

f (x,y) = ln |x|− 12

ln∣∣∣∣x+ 1

2y∣∣∣∣+g(y)

Then differentiating with respect to y,

− 12(2x+ y)

+g′ (y) =x

y(2x+ y)

g′ (y) =12y

and so g(y) = 12 ln |y| will work. Thus the integral curves are of the form

ln |x|− 12

ln∣∣∣∣x+ 1

2y∣∣∣∣+ 1

2ln |y|=C

You could simplify this if desired.

PROCEDURE 29.6.5 To solve

M (x,y)dx+N (x,y)dy = 0

using an integrating factor, do the following:

1. Look for an integrating factor µ which is a function of x alone. You do this if

My−Nx

Ndoes not depend on y. In this case, you solve

µ′ (x) = µ (x)

(My−Nx

)which is a separable equation. Solve and choose constant to satisfy initial condition.If this doesn’t work,

2. Look for an integrating factor µ which is a function of y alone. You do this if

Nx−My

Mdoes not depend on x. In this case, you solve

µ′ (y) =

Nx−My

Mµ (y)

which is a separable equation. Solve and choose constant to satisfy initial condition.

3. If neither of these work, check to see if M,N are both homogeneous of the samedegree. If they are, you could use either the methods of homogeneous equations orEuler’s formula for the integrating factor

1xM+ yN

4. If none of the above works, give up. You don’t know how to do it. The integratingfactor exists, but you don’t know how to find it.

29.6. THE INTEGRATING FACTOR 561Then integrating the first term with respect to x, the scalar potential is of the form1=! —<=lf (x,y) n |x| 2 n 21stp] +80)Then differentiating with respect to y,1 } xspay 78 Y= Ty2(2x+y) y(2x+y)i=xand so g(y) = $n |y| will work. Thus the integral curves are of the formg (y)]In|x|— >In 51 1ve by] + gid =CYou could simplify this if desired.PROCEDURE 29.6.5 1 solveM (x,y)dx+N (x,y)dy =0using an integrating factor, do the following:1. Look for an integrating factor which is a function of x alone. You do this ifMy, —N,Ndoes not depend on y. In this case, you solveH's) =n) (“*)which is a separable equation. Solve and choose constant to satisfy initial condition.If this doesn’t work,2. Look for an integrating factor & which is a function of y alone. You do this ifN,—MyMdoes not depend on x. In this case, you solveN,—My—y + (y)which is a separable equation. Solve and choose constant to satisfy initial condition.L(y) =3. If neither of these work, check to see if M,N are both homogeneous of the samedegree. If they are, you could use either the methods of homogeneous equations orEuler’s formula for the integrating factor1xM + yN°4. If none of the above works, give up. You don’t know how to do it. The integratingfactor exists, but you don’t know how to find it.