
1138 SECOND-ORDER LINEAR PDES: CLASSIFICATION, PROBLEMS, PARTICULAR SOLUTIONS
then k = 0; in this case with a 6= 0, the general solution of the corresponding hyperbolic
equation has the form
w(x, y) = exp(β
1
ξ + β
2
η)
f(ξ) + g(η)
, D = b
2
−ac > 0,
ξ = ay −
b +
√
D
x, η = ay −
b −
√
D
x,
β
1
=
aq −bp
4aD
+
p
4a
√
D
, β
2
=
aq − bp
4aD
−
p
4a
√
D
,
where f(ξ) and g(ξ) are arbitrary functions.
3
◦
. In the degenerate case b
2
−ac = 0, aq −bp = 0 (where the original equation is reduced
to an ordinary differential equation; see the last row in Table 14.1), the general solution of
Eq. (14.1.1.10) is expressed as
w = exp
−
px
2a
f(ay −bx) exp
x
√
λ
2a
+g(ay −bx) exp
−
x
√
λ
2a
if λ= p
2
−4as > 0