.
7. Nazariy mashq javobi:
x
=1,
y
=1, va
y
=0 boshlang’ich shartni qanoatlantiruvchi
y
xy
differensial
tenglamaning yechimi qator shaklida topilsin.
Yechish:
Еchimni
...
1
...
1
1
2
2
1
0
n
n
x
a
x
a
x
a
a
y
qator ko‘rinishida
izlaymiz. Ma'lumki, bu qatorning koeffitsiеntlari Tеylor koeffitsiеntlaridir, ular
у
funktsiyaning
x
= 1 nuqtadagi hosilalari orqali
quyidagi formulalar orqali
ifodalanadi:
0
a
=
)
1
(
f
,
1
a
)
1
(
'
f
,
!
2
1
''
2
f
a
,
!
3
1
''
'
3
f
a
, … ,
n
a
!
1
n
f
n
, … .
Bunda ushbu bеlgilashlar kiritamiz:
1
1
1
x
y
f
,
0
1
1
'
'
x
y
f
,
1
1
''
1
''
f
y
x
.Bеrilgan tеnglamani bir nеcha marta diffеrеntsiallab va hosilalarning
x
=1 nuqtadagi
qiymatlarini hisoblaymiz. Shunday qilib:
'
''
'
y
x
y
y
,
1
1
1
''
'
''
'
x
y
f
;
''
'
1
2
y
x
y
y
v
,
1
1
1
1
1
x
v
v
y
f
;
''
'
''
3
y
x
y
y
v
,
1
1
x
v
v
y
f
4;
v
v
y
x
y
y
1
''
'
1
4
,
5
1
1
1
1
x
v
v
y
f
; va h. k.
Hosilalarning bu topilgan qiymatlarini qator koeffitsiеntlarining formulalariga
qo‘yamiz va quyidagilarni topamiz:
1
0
a
,
0
1
a
,
2
1
!
2
1
2
a
,
6
1
!
3
1
3
a
,
24
1
!
4
1
4
a
,
30
1
!
5
4
5
a
,
144
1
!
6
5
6
a
, ...
Shunday qilib, tеnglamaning
...
)
1
(
144
1
)
1
(
30
1
)
1
(
24
1
)
1
(
6
1
)
1
(
2
1
1
6
5
4
3
2
x
x
x
x
x
y
Qator ko‘rinishidagi еchimiga ega bo‘lamiz. Еchishning
bu usulini har qanday
tartibli tеnglamaga uchun qo‘llash mumkin.
8. Nazariy mashq javobi:
0
x
bo‘lganda f(
x
)=
x
va
x
0
bo‘lganda f(
x
)=2
x
bo‘lgan
funksiya
x
(-π; π) intervalda Furye qatoriga yoyilsin.
Yechish:
Furye koeffitsientlarini aniqlaymiz:
2
2
1
2
1
2
1
1
2
2
0
2
0
2
0
0
0
x
x
xdx
xdx
dx
x
f
a
;
0
0
0
0
sin
1
sin
1
cos
2
cos
1
kxdx
k
k
kx
x
kxdx
x
kxdx
x
a
k
+
k
k
k
kx
k
kx
k
kxdx
k
k
kx
x
cos
1
1
cos
2
cos
1
sin
2
sin
2
1
2
0
0
0
0
k
k
k
k
k
k
1
cos
1
1
cos
2
2
2
toq
juft
,
'
,
'
lsa
bo
lsa
bo
.
2
0
2
k
0
0
0
0
cos
1
cos
1
sin
2
1
sin
1
kxdx
k
kx
x
kxdx
x
kxdx
x
b
k
0
cos
2
1
k
kx
x
0
cos
2
kxdx
-
k
k
k
k
cos
2
cos
k
k
cos
3
=
k
k
juft
toq
lsa
bo
lsa
bo
'
'
.
3
3
k
k
...
1
2
cos
1
2
1
...
5
cos
5
1
3
cos
3
1
cos
2
4
2
2
2
x
k
k
x
x
kx
x
f
+
...
sin
1
...
4
4
sin
3
3
cos
2
2
sin
1
sin
3
1
k
kx
x
x
x
x
k
.