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

restart;

with(DEtools):


L := an( x ) DF
( n )
a1( x ) DF  a0( x )
ai( x )
va
D DF( x )  1
DF( u v )  u DF( v )  DF( u ) v .

Differensial operatorlar bilan algebraik amallar bajarish. Differensial operatorlarni qo'shish, ko'paytirish, o'rniga qo'yish va boshqalarga o'xshash o'rin almashtirib bo'lmaydi (noncommutative domain) . C(x)[DF]dagi L

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

( n )
differensial operatori bu + . . . + a1( x ) DF  a0( x ) , bo'lib , bu yerda C(x)ning
L := an( x ) DF ai( x )
elementlari. C(x)[Dx] dagi L elementi L( y(x) )=0 chiziqli bir xil differensial tenglamaga to'g'ri keladi.
C(x)[Dx] doirasida ko'paytirish differensial operatorlarni shakllantirishga to'g'ri keladi. Shunday qilib agar L = mult(f,g) bo'lsa, u holda L( y(x) ) = f(g( y(x) )). Xususan, mult(DF,x) = x*DF + 1.
Misol tariqasida qanday algebraik amallarni bajarish mumkinligini ko'rish uchun quyidagi differensial operatorlarga e'tibor bering:

• 
  L1 := x^2*DF^2 - x*DF + (a-x^2);
  L1 := x2 DF2  x DF  a  x2
  
• 
  L2 := x*DF - (x^2-b);
  L2 := x DF  x2  b
  
• 
  L3 := DF^2 - x;
  L3 := DF2  x
  

Biz bu operatorlari ko'paytirib shuni eslatib o'tishimiz mumkinki, ularni o'rnini almashtirib bo'lmaydi: L4 := collect( mult( L1, L2, [DF, x] ), DF );

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

Argument [DF, x] mult komandasiga ko'paytirish DF va x belgilab bergan differensial sohasi ustida ekanligini ko'rsatadi, shuning uchun a va b o'zgaruvchilari o'zgarmas songa teng. Bu yana _Envdiffopdomain := [DF, x], muhit o'zrgaruvchisi orqali belgilanishi mumkin.
Bir tomonli eng kichik umumiy ko'paytuvchi va eng katta umumiy bo'luvchi tushunchasi shunday sohalarda mavjud bo'ladi. Misol tariqasida,

• 
  L6 := LCLM( L3, L2, [DF, x] );
  

L4 := x3 DF3  ( x4  x2  x2 b ) DF2  ( a x  x b  x  4 x3 ) DF  a x2  x4  x2 b  a b

• 
  L5 := collect( mult( L2, L1, [DF, x] ), DF );
  L5 :=
  x3 DF3  ( x4  x2  x2 b ) DF2  ( x b  x  a x ) DF  a b  x4  2 x2  a x2  x2 b
  

3
( b3  x3  2 b  3 x4 b  3 x2 b2  3 x4  3 b2  x3 b  x6  x5 ) DF2

x ( b  b2  x2  2 x2 b  x3  x4 )
 x DF
L6 := DF 
 b3  3 b  3 x4 b  3 x2 b2  2 x4  4 b2  x3 b  x6  x5  x2  2 x2 b
b  b2  x2  2 x2 b  x3  x4
Fan: Kompyuter algebrasi tizimlari
O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
va

• 
  L7 := GCRD( L4, L6, [DF, x] );
  

Buni o'ng yoki chap tomonni ko'paytirishi orqali tekshirish mumkin. Bizning masalamizda biz quyidagiga egamiz:

• 
  rightdivision(L6,L7, [DF,x] );
  

Ikki holatda ham yuqoridagi amal 0 qoldiqli bo'linmani beradi. Hisobni to'g'riligi quyidagi ko'paytirish orqali tekshirish mumkin:

• 
  collect( L4 - mult( %[1], L7, [DF,x] ), DF, normal );
  

L7 := DF 
x2  b
x
2
x ( b  b2  x2  2 x2 b  x3  x4 )

 ( x3  2 b  2 x4  2 b2 ) DF
 DF 
b  b2  x2  2 x2 b  x3  x4

• 
  rightdivision(L4,L7, [DF,x] );
  

 2 x3  4 x b  6 b  x4  2 x2  x b2  2 x3 b  x5



, 0 
[ x3 DF2  x2 DF  x3  a x  x, 0 ]
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov II-kurs