7-Mavzu: Mapleda differensial tenglamalarni yechish. Differensial tenglamalarni yechish funksiyalari. 1-, 2- va yuqori tartibli differensial tenglamalarni yechish

Qo'shimcha faktorlarga ajratish
Differensial operator

• df := DF^2+21/100*(x^2-x+1)/x^2/(x-1)^2;

Q(x)[DF]ning elementi sifatida faktorlarga ajralmasligi bilan ma'lum. Biz buni DFactor funksiyasi orqali tekshirishimiz mumkin:

• DFactor( df, [DF, x] );

yettinchi tartibning oltinchi simmetrik
Almerning ishlaridan kelib chiqishicha darajasi

• dff := symmetric_power(df,6,[DF,x]);

df := DF 
2
21 ( x2  x  1 )
100 x2 ( x  1 )2
 

 DF 
2
 21 ( x2  x  1 ) 
100 x2 ( x  1 )2
Bundan tashqari, uning differensial Galois maydoni A5 deb tanilgan. Zinger va
SL
2
7
294 ( x2  x  1 ) DF5 147 ( 2 x3  3 x2  5 x  2 ) DF4
25 x2 ( x  1 )2 5 x3 ( x  1 )3

dff := DF 
 441 ( 349 x4  698 x3  1547 x2  1198 x  349 ) DF3
441 ( 4829 x  8470 x2  2735 x4  7470 x3  1094 x5  1094 ) DF2
625 x5 ( x  1 )5

 63 (
8620804 x  19286733 x2  15457733 x4  22929126 x3  4791804 x5
625 x4 ( x  1 )4  1597268 x6  1597268 ) DF ( 62500 x6 ( x  1 )6 )  189 ( 1768603 x
 4817148 x2  6563550 x4  7263809 x3  3595389 x5  969969 x6  277134 x7
 277134 ) ( 31250 x7 ( x  1 )7 )
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari
faktorlar va faktorlar
O’qituvchi: T.Djiyanov II-kurs
3-chi va 4-chi tartibda . Lekin,
ko'paytuvchilarning
darajalari qanaqaligi ma'lum bo'lganligiga qaramasdan avvalgi algoritmlar ko'paytuvchilarga ajratishni amalga oshira olmagan edi. Biz quyidagiga ega bo'lamiz (biroz vaqtdan so'ng)

• fact_df := DFactor( dff, [DF,x] );

Faktorizasiyani tekshirish mumkin:

• L := mult( fact_df[1], fact_df[2], [DF,x]):
• collect( L - dff, DF, normal );

4
( x  1 ) x
5 ( 1  2 x  x2 ) x2

 4 ( 2 x  1 ) DF3 3 ( 23 x2  23 x  3 ) DF2
fact_df :=  DF 
50 ( x  1 ) x3 ( 1  2 x  x2 )
96 ( x4  2 x3  3 x2  2 x  1 )
625 ( x  1 ) ( 
1  3 x  3 x  x ) x
2 3 4
 3 ( 58 x3  87 x2  33 x  2 ) DF 
3
, DF
4 ( 2 x  1 ) DF2 ( 749 x2  749 x  249 ) DF
( x  1 ) x
25 ( 1  2 x  x2 ) x2
 

50 x3 ( 1  3 x  3 x2  x3 )

 9 ( 399 x2  265 x  66  266 x3 ) 

Fan: Kompyuter algebrasi tizimlari
7-Mavzu..
O’qituvchi: T.Djiyanov II-kurs
1. Differensial tenglamalarni qurish
ma'lum
ega differensial operatorlar. Haqiqatdan biz
Bessel funksiyasi yechimlariga quyidagiga egamiz:

• diffop2de(df,y(x),[DF,x]);

darajasi
yuqoridagining
darajaga
ko'tarilgan
differensial tenglamalarni hosil qiladi. Misol
df ning uchinchi simmetrik kombinasiyalari javobi bo'lgan tariqasida,

• dff := symmetric_power(df,3,[DF,x]);

• diffop2de(dff,y(x),[DF,x]):
• dsolve( % , y(x) );

DEtools da differensial operatorlar imkoniyatlaridan foydalanib tenglamalardan differerensial tenglamalarni hosil qilish mumkin. Masalan,

• df := x^2*DF^2 - x*DF + (x^2 - 1);

df := x2 DF2  x DF  x2  1

2



 d
( x  1 ) y( x )  x  dx y( x )   x





2  d2
dx
2
y( x )

• dsolve(%,y(x));

y( x)  _C1 x BesselJ( 2, x)  _C2 x BesselY( 2, x)

 

dff := DF 
4
x
6 DF3 5 ( 2 x2  1 ) DF2 5 ( 6 x2  7 ) DF 3 ( 2 x2  21  3 x4 )
x2 x3 x4
Fan: Kompyuter algebrasi tizimlari
O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
Biz symmetric_product funksiyasidan foydalanib javoblarning ko'paytmalarini qurishimiz mumkin. Misol uchun,

• df2 := DF - x:
• dsolve( diffop2de(df2,y(x),[DF,x]), y(x) );

df va df2 ning birgalikdagi yechimining ko'paytmasini beradi.
Eksponensial natijaga ega bo'lgan tanglamaning yaratish uchun biz to'g'ri operatorlarni eksponensial tenglamalardan foydalanib hosil
to'g'ridan- qilishimiz
mumkin yoki tashqi darajaga ko'tarish operasiyasidan foydalanishimiz mumkin.
2 , x )3  _C2 x3 BesselY( 2 , x )3
2 , x ) BesselY( 2 , x )2
2 , x )2 BesselY( 2 , x )
y( x )  _C1 x3 BesselJ(

• _C3 x3 BesselJ(
• _C4 x3 BesselJ(


 x2 
 2 
y( x )  _C1 e
eksponensial natijani hosil qiladi. Bu o'rinda

• df3 := symmetric_product( df, df2, [DF,x] );

df3 := x2 DF2  x DF  2 x3 DF  x2  x4  1

• dsolve( diffop2de( df3,y(x),[DF,x]), y(x) );

 x2   x2 
 2   2 
y( x )  _C1 e  x BesselJ( 2 , x )  _C2 e  x BesselY( 2 , x )

Agar operator L to'g'ri darajali n ko'paytmasiga ega bo'lsa u holda L ning n-chi tashqi darajasi 1-chi darajali to'g'ri ko'paytuvchiga ega bo'ladi. Misol uchun L := collect( mult( df+x, df, [DF,x]), DF );

Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs 7-Mavzu..
L := x4 DF4  2 x3 DF3  ( 3 x2  2 x4  x3 ) DF2  ( 2 x3  3 x  x2 ) DF  x4  1  x3
 x  2 x2

• M := exterior_power(L,2,[DF,x]);

6
x
8 DF5 ( 7  4 x2  2 x ) DF4 ( 7 x  3  24 x2 ) DF3
x2 x3
 
M := DF 
 ( 25 x2  3 x  7 ) DF2  ( x2  14 ) DF  x2  14
x4 x5 x6

• DFactor(M, [DF,x]);

 
 
 DF  x , DF 
4
x
 4 5 DF3 2 ( 2  x  2 x2 ) DF2
x2
3 ( 1  4 x ) DF 5
x2
x
2

• , DF 

1 
x 
O'ng tomondagi to'g'ri ko'paytuvchini ko'rsatadi. Bu quyidagi orqali ko'rinishi mumkin

• expsols( diffop2de(M,y(x),[DF,x]), y(x) );

[ x ]

5.Chiziqli differensial tenglamalarning yopiq ko`rinishdagi yechimlari

Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs 7-Mavzu..

Maple ning dsolve komandasi chiziqli differensial tengamalarning yopiq ko'rinishdagi yechimlari topish imkonini beradi.

Ikkinchi tartibdagi tenglamalar

dsolve komandasi quyidagi shakldagi homogeneous ikkinchi tartibdagi differensial tenglamani o'zgartiradi.

quyidagi shaklga

tenglamaning yechimlarini qurishda foydalanadi. Bu usul ikkinchi tenglamaga yechimni I(x)ning qisman ko'paytmalari kengaytirish orqali hisoblaydi.
Misol uchun, bizda (Kemke tomonidan klassik tekstda berilishicha) mavjud:

• restart;
• ode1 := diff(y(x),x$2) + 2*x^2*diff(y(x),x) + 2*x*y(x);



2
dx
2

   dx
ode := a( x )  d y( x )   b( x )  d y( x )   c( x ) y( x )  0
new_ode :=

 
 

 d2
dx
2
y( x ) I( x ) y( x )  0 , keyin ikkinchi tenglamaning yechimlarini birinchi
ode1 :=




dx
2



 d2  2  d
y( x )  2 x  dx y( x )   2 x y( x )
Fan: Kompyuter algebrasi tizimlari
O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..

• ans1 := dsolve(ode1, y(x) );

• ode2 := (4*x^4-1)*diff(diff(y(x),x),x)+(12*x^4+1)/x*diff(y(x),x)+16*x^2*(-u^2/(4*x^4-1)- v*(v+1))*y(x);


  x3 
 3 
 1 x3 
ans1 := y( x )  _C1 e x BesselI ,   _C2 e

  x3 
 3 

x BesselK ,
 1 x3 
 6 3   6 3 
4




 d2 
dx
2

 dx
( 12 x4  1 )  d y( x ) 
x
y( x ) 
ode2 := ( 4 x  1 )
2 

 


u2
4
4 x  1
 v ( v  1 ) y( x )
 16 x 

• ans2 := dsolve(ode2, y(x) );

ans2 := y( x )  _C1 LegendreP ( v, u, 2 x2 )  _C2 LegendreQ ( v, u, 2 x2 )

• ode3 := diff(y(x),x$2) - x*diff(y(x),x) + n*y(x);

ode3 :=




dx
2





 d2   d
dx
y( x )  x y( x )  n y( x )

• ans3 := dsolve( ode3, y(x) );

ans3 := y( x ) 

 4 
_C1 e  WhittakerM  , ,
 1 n 1 x2 
 x2   x2 
 4 
 
_C2 e  WhittakerW  , ,
 1 n 1 x2 
 4 2 4 2    4 2 4 2 
x x

• ode4 := (x-9*x^5)*diff(y(x),x$2) + (1-9*x^4)*diff(y(x),x) + 36*x^3*y(x);

ode4 := ( x  9 x )
5




dx
2
4





 d2   d
dx
3
y( x )  ( 1  9 x ) y( x )  36 x y( x )
II-kurs
7-Mavzu..
odetest komandasidan foydalangan holda
Barcha quyidagi misollarda yechimlarni tekshirish mumkin:

• odetest( ans1, ode1 );

• odetest( ans2, ode2 );

• odetest( ans3, ode3 );

• odetest( ans4, ode4 );

Yuqori tartibdagi tenglamalar
Biz yuqori tartibdagi tenglamalarni ham yechishimiz mumkin. Misol uchun, quyidagi tenglamada
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov

• ans4 := dsolve( ode4, y(x) );

ans4 := y( x )  _C1 EllipticE( 3 x2 )  _C2 ( EllipticCE( 3 x2 )  EllipticCK( 3 x2 ) )

0
0
0
0
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs

• ode := -300*y(x) + 192*x*diff(y(x),x)+(-47*x^2+4*x^3*mu-x^4- 4*x^2*nu^2)*diff(y(x),x$2) +12*y(x)*x^2-

48*y(x)*mu*x+48*y(x)*nu^2+4*x^4*diff(y(x),x$4);
DEtools[ratsols] rasional funksiya yechuvchisidan foydalanib ikkita yechimini topishimiz mumkin. Tartibni qisqartirish ohirgi natijani integrallar ko'rinishida hosil qiladi:

• dsolve( ode, y(x) );


 
2 3 4 2 2





 d  d2
dx
2
y( x )
ode := 300 y( x )  192 x  dx y( x )   ( 47 x  4 x   x  4 x  )

 


4  d4
dx
4
y( x )
 12 y( x ) x2  48 y( x )  x  48 y( x ) 2  4 x
x3
y( x )  _C1  _C2 x4



 4
_C3  x WhittakerM( , , x ) dx  
 
 WhittakerM( , , x )
x3
dx x 

7 
x3




 4

_C4  x WhittakerW( , , x ) dx  

 WhittakerW( , , x )
x3
dx x 

7 
x3


• odetest( %, ode );

0
Maple tez test o'tkazib ma'lum ODE ikkinchi tartibdagi tenglamaning simmetrik ko'paytmasi ekanligini tekshirib ko'radi. Agar shunday bo'lsa, javob ikkinchi

tartibdagi tenglamaning yechimlaridan aniqlanishi mumkin. Misol uchun, quyidagi Kemke yoki Abramovits va Stegun da ko'rish mumkin bo'lgan tenglama ode := diff(y(x),x$3) - 4*x*diff(y(x),x) - 2*y(x);

Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs 7-Mavzu..
Xuddi shunday q'xshash tarzda,

• ode2 := diff(y(x),x$3)-6*x*diff(y(x),x$2) +2*(4*x^2+2*a-1)*diff(y(x),x)-8*a*x*y(x);

• ans2 := dsolve( ode2, y(x) );

• odetest(ans2, ode2 );

ode :=

 
3

dx  
 
 d3   d
 dx
y( x )  4 x y( x )  2 y( x )
Eyri tenglamasining simmetrik darajasiga tengdir. Shunday qilib, dsolve quyidagini hosil qiladi:

• dsolve( ode, y(x) );

y( x )  _C1 AiryBi( x )2  _C2 AiryAi( x )2  _C3 AiryAi( x ) AiryBi( x )

ode2 :=

3
   
  
dx dx
2
2



 d3   d2   d
y( x )  6 x y( x )  2 ( 4 x  2 a  1 )  dx y( x )   8 a x y( x )
( x2 )
 4 
1 a 1
4 4
_C1 e WhittakerM  , , x2 
2
x
ans2 := y( x ) 
( x2 )

 4
1 a 1
4 4
_C2 e WhittakerW  , , x2 
2
x

 4 
 4 
4 4 4 4
( x2 )  1 a 1  1 a 1
_C3 e WhittakerM  , , x2  WhittakerW  , , x2 
x

Fan: Kompyuter algebrasi tizimlari
O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
Yuqoridagi metodlar bilan birgalikda, quyidagi tenglama,

• ode := 2*y(x)+(-2+4*x)*diff(y(x),x)-4*x*diff(y(x),x$2)

-diff(y(x),x$3)+diff(y(x),x$4);

Maplening DEtools[expsols] eksponensial yechuvchisi orqali aniqlangan yagona yechimga ega ekanligini ko'rish mumkin, keyinchalik tartibni qisqartirish simmetrik tenglamaga olib keladi. Bu quyidagini beradi

• dsolve( ode, y(x) );

Chiziqli differensial tenglamalarning yechishdagi yaxshilanishlar tabiiy ravishda boshqa differensial tenglamalarni yechishda yaxshilanishlarga yo'l ochib beradi.
0





dx
ode := 2 y( x )  ( 2  4 x ) y( x )  4 x


 d  d2
dx
2
y( x ) 
 
 
dx
3
 

  
  d3   d4 
dx
4
y( x )  y( x )
AiryBi( x )2 e( x ) dx ex

y( x )  _C1 ex  _C2 AiryAi( x )2 e( x ) dx ex  _C3


• _C4 AiryAi( x ) AiryBi( x ) e( x ) dx ex



• odetest( %, ode );

0

Masalan, birinchi tartibda Rikatti tenglamalari yechimlarini qurish uchun odatda ikkinchi tartibdagi tenglamalarga keltiriladi. Shu tarzda, birinchi tartibdagi chiziqli sistemalari bitta yoki ko'proq yuqori tartibdagi skalyar tenglamalarni qurish orqali yechiladi va matrisa yechimlarini qurish orqali yechiladi. Yagona quyidagi misol orqali topilishi mumkin, ode := diff(y(x),x) = x + y(x) + a*y(x)^2;

Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs 7-Mavzu..

• dsolve( ode, y(x) );

Yuqoridagi yechimda, AiryAi(1, .. ) va AiryBi(1,...) funksiyalari o'z o'rnida AiryAi va AiryBi funksiyalrining hosilalarini ko'rsatadi.
dx
ode := d y( x )  x  y( x )  a y( x )2

y( x )   

_C1 AiryBi 1,  1  4 a x 
( 2/3 )
4 a






( 2/3 )
4 a


 
 
( 2/3 )
4 a
1
( 2/3 )   1  4 a x   1  4 a x   2
a  _C1 AiryBi    AiryAi   

AiryAi  1 
4 a
(

a
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs
Differensial tenglamalarni echish. dsolve funksiyasi
Differensial tenglamalarni echish masalasi matematikaning asosiy masalalaridan hisoblanadi. Mapleda differensial tenglamalarni ham ahalitik ham sonli usullarda echish imkoniyatlari mavjud.
SHuningdek, yakka tartibdagi bitta differensial tenglamani va tenglamalar
funksiyasidan foydalaniladi.
sistemasini ham echish mumkin.
Differensial tenglamalarni echishda dsolve
Funksiyaning quyidagi ko’rinishlari mavlud:
dsolve(ODT) dsolve(ODT,y(x),tip) dsolve({ODT,BSH},y(x),tip)
Bunda: ODT – oddiy differensial tenglama yoki differensial tenglamalar sistemasi, BSH – Boshlang’ich shartni ifodalovchi ifoda, y(x) – bir o’zgaruvchili funksiya, tip – echilayotgan differensial tenglama echimi turini aniqlovchi parametr bo’lib u quyidagi kabi qiymatlar qabul qilishi mumkin.
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs 7-Mavzu..

• exact – jimlik qoidasiga mos analitik echimini topadi;
• system – differensial tenglamalar sistemasining echimini topadi;
• formal series – echimni kophad ko’rinishida chiqaradi;
• numeric – sonli usulda echadi.

Koshi masalasini echish uchun dsolve fuknsiyasi argumentiga boshlang’ich shartni ham kiritish lozim. Chegaraviy masalani echish uchun esa chegaraviy shartlarni kirittish lozim. Agar differensial tenglama boshlang’ich va chegaraviy shartlar bilan echilsa, _C1,_C2, … ko’rinishdagi o’zgarmaslar paydo bo’lmayadi.
Differensial tenglamadagi hosila belgisi diff funksiyasi yoki D operatopi orqali ko’rsatiladi.
Dastlab birinchi tartibli bitta oddiy differensial tenglama echishga doir misollar bilan tanishamiz:

• dsolve(diff(y(x),x)-a*x=0,y(x));

• dsolve(diff(y(x),x)-y(x)=exp(-x),y(x));

2
y( x )  1 a x2  _C1


• dsolve(diff(y(x),x)-y(x)=sin(x)*x,y(x));


2
y( x )    1 e ( 2 x )  _C1  e x
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7-Mavzu..

• yx0:=y(0)=0,y(1)=1;

y( x )   1 cos( x ) x  1 cos( x )  1 sin( x ) x  e x _C1
2 2 2

• dsolve(diff(y(x),x)-y(x)=cos(x)*x^2,y(x));

y( x )   1 cos( x ) x2  1 cos( x )  1 sin( x ) x2  sin( x ) x  1 sin( x )  e x _C1
2 2 2 2
Ikkinchi tartibli differensial tenglamalarni echish
Ikkinchi tartibli differensial tenglamalarni echish uchun ham dsolve va diff
funksiyalaridan foydalaniladi. Faqat yuqori tartibli hosilalar $ belgisi orqali beriladi.

• dsolve(diff(y(x),x$2)-diff(y(x),x)=sin(x),y(x));

y( x )   1 sin( x )  1 cos( x )  e x _C1  _C2
2 2

• dt:=m*diff(y(x),x$2)-k*diff(y(x),x);

dt := m






 2
x
2





 
x
y( x )  k y( x )
yx0 := y( 0 )  0, y( 1 )  1
Fan: Kompyuter algebrasi tizimlari
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7-Mavzu..

• dsolve({dt,yx0},y(x));

Agar ikkinchi misolga e’tibor bersangiz, differensial tenglamaning o’zi va boshlang’ich shartlar qandaydir o’zgaruvchiga qiymat sifatida berilib, so’ng ulardan tenglamani echish jarayonida parametr sifatida foydalanilmoqda. Bu imkoniyat yaxshi imkoniyat bo’lib, bitta dsolve funksiyasidan foydalanib bir nechta differensial tenglamalar va ularni ham har xil boshlang’ch shartlarda echish mumkin bo’ladi. Bu usuldan foydalanilganda _CN ko’rinishdagi ixtiyoriy ozgarmas ham ishtirok etmaydi.
Differensial tenglamalar sistemasini echish
Quyidagi misollarda ikkita differensial tenglamalar sistemasi ikki xil usulda echilgan. YA’ni qatopga yoyish yoli bilan va Laplas almashtirishlaridan foydalanish yoli bilan. Shuni ta’kidlash lozimki, differensial tenglama qatorga yoyish yoli bilan echilganda taqribiy yechim topiladi. SHuning uchun ikki usulda topilgan echimlar farq qilmoqda.
y( x )  
1
 k  
 m 
 k x 
 
e  m 
1  e  1  e 
 
 k 
 m 
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7-Mavzu..

• sys:=diff(y(x),x)=2*z(x)-y(x)-x,diff(z(x),x)=y(x);

• fens:={y(x),z(x)};

• dsolve({sys,y(0)=0,z(0)=1},fens);

• dsolve({sys,y(0)=0,z(0)=1},fens,series);

sys :=  y( x )  2 z( x )  y( x )  x,  z( x )  y( x )
x x
fens := { z( x ), y( x )}
{ z( x )  5 e ( 2 x )  1 e x  1  1 x, y( x )   5 e ( 2 x )  1 e x  1 } 12 3 4 2 6 3 2
{ y( x ) 
2 x  3 x2  7 x3  13 x4  9 x5  53 x6  107 x7 
x8 
71 61
13440 51840
x9  O( x10 ),
2 6 24 40 720 5040
z( x ) 
53 107 71
1  x2  1 x3  7 x4  13 x5  3 x6  x7  x8  x9  O( x10 ) }
2 24 120 80 5040 40320 120960

• dsolve({sys,y(0)=0,z(0)=1},fens,laplace);

{ z( x )  1 e x  5 e ( 2 x )  1 x  1, y( x )   5 e ( 2 x )  1 e x  1 } 3 12 2 4 6 3 2
Maplening differensial tenglamalarni echish bo’yicha juda katta imkoniyatlari bo’lsa ham bu degan so’z har qanday differensial tenglamani analitik usulda echa oladi degani emas. Bunday hollarda conli usulda echish mumkin bo’ladi.

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O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
Differensial tenglamalarni sonli usulda echish
Ko’pchilik chiziqli bo’lmagan differensial tenglamalar analitik echimga ega bo’lmaydi. Bundan tashqari ko’pincha analitik echimning o’zi shart ham bo’lmaydi. CHunki natijani grafik ko’rinishda tasvirlash talab etiladi. Bunday hollarda differensial tenglamalar sonli usullarda echiladi. Bu holda ham dsolve funksiyasidan foydalaniladi, faqat yuqorida qayd etilganidek, numeric yoki type=numeric parametrlari bilan birgalikda foydalaniladi.

• sys:=diff(y(x),x)=2*z(x)-y(x)-x,diff(z(x),x)=y(x);

• fens:={y(x),z(x)};

sys :=  y( x )  2 z( x )  y( x )  x,  z( x )  y( x )
x x
fens := { z( x ), y( x )}

• F:=dsolve({sys,y(0)=0,z(0)=1},fens,numeric);

F := proc(rkf45_x) ... end proc

• F(2);

[ x  2., y( x )  2.94775557620857454, z( x )  3.72064994303630758 ]

II-kurs
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov

• plots[odeplot](F,[x,z(x)],0..2.5,labels=[x,z],color=black);

Sinov savollari

• Differensial tenglamalar ustida qanday algebraik amallar bajarish mumkin?
• Differensial tenglamaning umumiy yechimi qanday aniqlanadi?
• Differensial tenglama uchun Koshi masalasini qanday yechiladi?
• Darajali qatorlarga yoyib taqribiy yechim qanday aniqlanadi?
• Differensial operatorlarning faktorizasiyasi qanday bajariladi?
• Differensial tenglamalarni hosil qilish jarayonini tushuntiring.
• Chiziqli differensial tenglamalarning yopiq ko`rinishdagi yechimi qanday aniqlanadi?
• Yuqori tartiblii differensial tenglamalarning yopiq ko`rinishdagi yechimi qanday aniqlanadi?