Обыкновенные дифференциальные уравнения высших порядков. Мухарлямов Р.К - 40 стр.

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(x
2
1)y
00
+ 4xy
0
+ 2y = 0.
y
1
= u
2
u
1
=
1
x + 1
.
y
y
1
0
= C
1
(x + 1)
2
exp
Z
4xdx
x
2
1
y
y
1
0
=
C
1
(x 1)
2
y =
C
1
x + 1
+
C
2
x 1
.
y =
C
1
(x)
x + 1
+
C
2
(x)
x 1
.
C
1
(x) C
2
(x)
C
0
1
(x)
x+1
+
C
0
2
(x)
x1
= 0,
C
0
1
(x)
(x+1)
2
C
0
2
(x)
(x1)
2
=
6x
x
2
1
.
C
0
1
= 3x(x + 1),
C
0
2
= 3x(x 1).
C
1
(x) = x
3
+
3
2
x
2
+ C
1
,
C
2
(x) = x
3
+
3
2
x
2
+ C
2
.
C
1
(x) C
2
(x)
y =
C
1
x + 1
+
C
2
x 1
+
x
3
x
2
1
.
xy
00
+ 2y
0
xy = 0, y
1
= e
x
/x
y
00
2(1 + tg
2
x)y = 0, y
1
= tg x
xy
000
y
00
xy
0
+ y = 0, y
1
= x, y
2
= e
x
x
2
(2x 1)y
000
+ (4x 3)xy
00
2xy
0
+ 2y = 0, y
1
= x, y
2
= 1/x
(3x
3
+ x)y
00
+ 2y
0
6xy = 4 12x
2
, y
1
= 2x, y
2
= (x + 1)
2
xy = C
1
e
x
+ C
2
e
x
y = C
1
tg x + C
2
(1 + x) tg x
y = C
1
x + C
2
e
x
+ C
3
e
x
y = C
1
x + C
2
x
1
+ C
3
(x ln |x| + 1)
y = C
1
(x
2
+ 1) + C
2
x
1
+ 2x
                                                        40


   Ðåøåíèå.         Íàéäåì îáùåå ðåøåíèå ñîîòâåòñòâóþùåãî îäíîðîäíîãî óðàâíåíèÿ

                                         (x2 − 1)y 00 + 4xy 0 + 2y = 0.           (3.10)

Ó÷èòûâàÿ çàìå÷àíèå (1), íàõîäèì ÷àñòíîå ðåøåíèå óðàâíåíèÿ (3.10):
                                                               1
                                            y 1 = u2 − u1 =       .
                                                              x+1
Èñïîëüçóÿ ôîðìóëó Îñòðîãðàäñêîãî - Ëèóâèëëÿ (3.3), çàïèøåì îáùåå ðåøåíèå óðàâíåíèÿ
(3.10):
              0                    Z          0
          y                   2         4xdx      y        C1         C1   C2
                    = C1 (x + 1) exp −   2
                                               ⇒      =        2
                                                                 ⇒y=     +    .
          y1                            x −1      y1    (x − 1)      x+1 x−1
   Äëÿ íàõîæäåíèÿ îáùåãî ðåøåíèÿ íåîäíîðîäíîãî óðàâíåíèÿ (3.9) èñïîëüçóåì ìåòîä
âàðèàöèè ïîñòîÿííûõ. Ðåøåíèå èùåì â âèäå
                                                  C1 (x) C2 (x)
                                             y=         +       .                 (3.11)
                                                  x+1     x−1
Ôóíêöèè C1 (x) è C2 (x) îïðåäåëèì èç ñèñòåìû
                                                       
    C10 (x) + C20 (x) = 0,          C 0 = 3x(x + 1),    C (x) = x3 + 3 x2 + C ,
       x+1     x−1                     1                    1           2       1
           0          0
                                  ⇒                    ⇒
    − C1 (x) − C2 (x) = 6x .        C 0 = 3x(x − 1).    C (x) = −x3 + 3 x2 + C .
        (x+1)2      (x−1)2  x2 −1      2                    2              2       2


   Ïîäñòàâèì C1 (x) è C2 (x) â ôîðìóëó (3.11) è ïîëó÷èì èñêîìîå ðåøåíèå:

                                              C1   C2    x3
                                        y=       +    + 2   .
                                             x+1 x−1 x −1
   Íàéòè ðåøåíèÿ óðàâíåíèé:

   56.    xy 00 + 2y 0 − xy = 0, y1 = ex /x.
   57.    y 00 − 2(1 + tg2 x)y = 0, y1 = tg x.
   58.    xy 000 − y 00 − xy 0 + y = 0, y1 = x, y2 = ex .
   59.    x2 (2x − 1)y 000 + (4x − 3)xy 00 − 2xy 0 + 2y = 0, y1 = x, y2 = 1/x.
   60.    (3x3 + x)y 00 + 2y 0 − 6xy = 4 − 12x2 , y1 = 2x, y2 = (x + 1)2 .
   Îòâåòû:

   56.    xy = C1 e−x + C2 ex .
   57.    y = C1 tg x + C2 (1 + x) tg x.
   58.    y = C1 x + C2 ex + C3 e−x .
   59.    y = C1 x + C2 x−1 + C3 (x ln |x| + 1).
   60.    y = C1 (x2 + 1) + C2 x−1 + 2x.