5.2. MONGE–AMPERE EQUATION 103
where ˛ D x C p, ˇ D u C
1
2
p
2
, and D 2x q. With these variables, F is independent of x;
this leads to
F
ˇ
D 0; F
y
C F
˛
C F
D 0; (5.86)
from which we obtain the final form of first integral
F .x y Cp; y 2x C q/ D 0: (5.87)
Any choice of F in (5.82) or (5.87) will lead to an exact solution of the original PDE. For
example, from (5.87), choosing
a.x y C p/ C b.y 2x C q/ D 0; (5.88)
where a and b are arbitrary constants and f is an arbitrary function, gives
u D
b
2
ab a
2
2a
2
x
2
C
2a b
a
xy Cf .bx ay/; (5.89)
an exact solution to (5.76).
Example 5.6 Obtain first integral to
xqr .x Cy/s Cypt C xy
rt s
2
D 1 pq: (5.90)
Identifying that A D xq, B D .x C y/, C D yp, D D xy, and E D 1 pq, then from (5.72)
gives
x
2
y
2
2
C xy.x C y/ C xypq C xy.1 pq/ D 0; (5.91)
leading to D
1
x
;
1
y
. us, we have two cases to consider.
Case 1: From (5.74) we have
xF
x
C xpF
u
pF
p
F
q
D 0; (5.92a)
yF
y
C yqF
u
F
p
qF
q
D 0: (5.92b)
Using the method of characteristic, we solve the first, giving
F D F
.
xp; y; u xp ln x; ln x Cq
/
: (5.93)
Substituting into the second gives
F
y
C y. ln x/F
ˇ
xF
˛
C x ln xF
ˇ
C
.
ln x
/
F
D 0; (5.94)
or, re-arranging gives
F
y
C yF
ˇ
F
xF
˛
C ln x
F
yF
ˇ
C x ln xF
ˇ
D 0; (5.95)