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y2
?(?+3/2)
D + (2? + 1)M, T (? ? R) : v = y1 ?(?), ?= ,
y1
3 5
(? 2 + 1)? + (5 + 2?)? ? +
? ? +? + ? ? = 0.
2 2
316 P. Basarab-Horwath, L. Barannyk, W.I. Fushchych

For ? = ? 5 we have
2

? = C1 ? + C2 .
If ? = ? 3 then
2

? = C1 arctan ? + C2 .
For ? = ? 3 , ? 5 then
2 2

3
? = C1 (1 + ? 2 )?(?/2+3/4) cos + ? arctan ? + C2 .
2
The exact solutions being:
5
?=? ,
v = C1 y2 + C2 y2 ,
2
y2 3
?=? ,
v = C1 arctan + C2 ,
y1 2
3 y2 35
v = C1 (y1 + y2 )?(?/2+3/4) cos ? = ? ,? .
2 2
+ ? arctan + C2
2 y1 22
3.1.3.
2
y1
?(?+3/4)
D + (4? + 1)M, P2 (? ? R) : v=t ?(?), ?= ,
t
3
4? ? + (2 + ?)? +
? ? + ? ? = 0.
4
If we make the transformation ? > ? = ? ? in this ODE, we obtain
4

1 3
?? ? ? ?+
?? + ? = 0,
2 4
where ? denotes differentiation with respect to ?. The solutions of this equation are
given in terms of the Pochhammer–Barnes confluent hypergeometric function (see
for example Vol. 1, ch. 6 of [11])
?
(a)n z n
?(a; b; z) =
(b)n n!
n=0

with b = 0 and where (a)n = a(a + 1)(a + 2) · · · (a + n ? 1), n ? 1. We find then [11]
1/2
31 1 1 53 1
? = C1 ? ? + ; ; ? ? + C2 ? ? ? ? + ; ;? ? .
42 4 4 42 4
Thus we find the exact solution
1/2
2 2 2
3 1 y1 y1 5 3 y1
?(?+ 3 )
C1 ? ? + ; ? ? ? ? + ; ;?
v=t + C2 .
4
42 4t 4t 42 4t

3.1.4.
2
y1 1
v = exp ?
G1 , P2 : ?(?), ? = t, ?+
? ?=0
4t 2?
New solutions of the wave equation 317

which integrates to give the exact solution
2
y1
?1/2
exp ?
v = C|t| .
4t
3.1.5.
t3 y1 t
? ? = t2 ? 2y1 ,
P2 , T + G1 : v = exp ?(?),
6 2
16? ? ?? = 0.
?
v
??(z) with z = ? 3/2 /6. Then ? satisfies
To treat this ODE, first write ? =
1 1
? + ? ? 1+ 2 ?=0
z 9z
which is the equation for the Bessel function J±1/3 (iz) (these two are linearly
independent solutions) (see Vol. 2, section 7.2.2 of [11]). Consequently, we have
t3 ty1
1/2
v = t2 ? 2y1 ? ?
exp
6 2
3/2 3/2
i t2 ? 2y1 i t2 ? 2y1
? C1 J1/3 + C2 J?1/3
6 6

as an exact solution of the heat equation.
3.1.6.
J12 + ?D ? ?(4? + 2)M, T (? > 0, ? ? R),
y1 1
2?
2 2 2
v = y1 + y2 ?(?), ? = ? arctan + ln y1 + y2 ,
y2 2
2 2
(? + 1)? + 4? ? + 4? ? = 0.
? ?

Integrating this equation, we obtain

for ? = 0
? = C1 ? + C2

and
2?? 2???
? = C1 exp ? cos ? for ? = 0.
+ C2
1 + ?2 1 + ?2
These then give us the exact solutions
y1
1 2 2
ln y1 + y2 + C2 for ? = 0,
v = C1 ? arctan +
y2
2
2?? 2???
2?
y1 + y2 exp ?
2
for ? = 0,
v = C1 cos + C2
1 + ?2 1 + ?2
where
y1 1 2 2
? = ? arctan + ln y1 + y2 .
y2 2
318 P. Basarab-Horwath, L. Barannyk, W.I. Fushchych

3.1.7.

J12 + 2?M, D ? (4? + 2)M (? ? 0, ? ? R),
2 2
y1 y1 + y 2
?
v = t exp ? arctan ?(?), ? = ,
y2 t
?2 ?2 ??
?
2
? ?+ ?+
? ?+
? ? = 0.
4 4 4

This equation gives

?2
1 1 ?
?
?+
? + ?+
? ? = 0.
4? 2
4? 4?

Its solutions can be given in terms of Whittaker functions W (k; m; z) (see Vol. 1,
ch. 6, pp. 248–251 of [11]) and one obtains

e??/8 i? ?
?= v W ?(? + 1/2); ;
24
?

and hence

t?+1/2 2 2 2 2
y1 + y2 y1 i? y1 + y2
exp ? W ?(? + 1/2); ;
v= exp ? arctan .
8t y2 2 4t
2 2
y1 + y2

3.1.8.

J12 + 2?M, T + ?M (? ? 0, ? = 0, ±1),
y1 ?t
? 2 2
v = exp ? arctan ?(?), ? = y1 + y2 ,
y2 2
2
? ??
?2 ? + ?? +
? ? + ? = 0.
4 8

We have the following cases:

? = C1 + C2 log ? for ? = ? = 0,
?
? = C1 cos ? log ? + C2 for ? = 0, ? = 0,
2
??
for ? ? 0, ? = 0.
? = Ji?
2

Consequently, we have the following solutions of (8)
2 2
for ? = ? = 0,
v = C1 + C2 log y1 + y2
y1 ?
C1 cos ? log y1 + y2 + C2
2 2
for ? = 0, ? = 0,
v = exp ? arctan
y2 2
2 2
y1 ?t ? (y1 + y2 )
? for ? ? 0, ? = 0.
v = exp ? arctan Ji?
y2 2 2
New solutions of the wave equation 319

3.1.9.
(? ? R),
J12 + S + T + 2?M, G1 + P2
2
1 ? t2 2
y1 + ty2 y1
?1/2
? ? ? arctan t ?(?),
2
v = t +1 exp
t2 + 1
4t 4t
y1 + ty2
? + (? + ? 2 )? = 0.
?= , ?
t2 + 1
This equation is known as the Weber equation. Its solutions are the real and imaginary
parts of the functions
D?v? (±(1 + i)?),
where D? (z) are the Weber–Hermite (parabolic cylinder) functions (Vol. 2, ch. 8,
section 8.2 of [11]). This gives the following exact solutions of (9):
2

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