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

y(0):=1: D(y)(0):=0:f;

y(x)  1  x2  1 x3  O(x4 )

6
Koshi masalasining to 6-chi tartibli darajali qator ko’rinishida yaqinlashuvchi
yechimini hamda aniq yechimlarini topamiz: y  y  3(2  x2 ) sin x , y(0)  1 , y(0)  1, y(0)  1 . Aniq hamda yaqinlashuvchi yechimlarining grafigini solishtirish uchun bitta rasmda chizamiz.

• 
  restart; Order:=6:
  
• 
  de:=diff(y(x),x$3)-diff(y(x),x)= 3*(2-x^2)*sin(x);
  

2
3



 x
 
  y( x)  3(2  x ) sin(x)




de:= y( x)

 x
 3
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
2 2 4 4

• 
  cond:=y(0)=1, D(y)(0)=1, (D@@2)(y)(0)=1;
  cond:=y(0)=1, D(y)(0)=1, D(2)(y)(0)=1
  
• 
  dsolve({de,cond},y(x));
  y(x)= 21 cos( x)  3 x2 cos( x)  6x sin(x)  12  7 ex  3 e(x)
  

2 6 24 120

• 
  y1:=rhs(%):
  
• 
  dsolve({de,cond},y(x), series);
  y(x)=1  x  1 x2  1 x3  7 x4  1 x5  O(x6 )
  

Differensial operatorlarning faktorizasiyasi
Bu bo'limda yopiq holatdagi yechimlar va boshqa faktorizasiya opsiyalariga ta'luqli bo'limlar mavjud.
Differensial tenglamalarning yopiq shakldagi yechimlarini topish quyidagi uchinchi tartibdagi differensial tenlamaga e'tibor qarating

• 
  ode := (-x-x^3)*y(x)-2*x^2*diff(y(x),x)+(x^3+x)*diff(y(x),x$2)+x^2*diff(y(x),x$3);
  

3



 3
ode := ( x  x ) y( x )  2 x  dx y( x )   ( x  x )



2  d  d2
dx
2
y( x )  x

 


 2  d3 
dx
3
y( x )
II-kurs
7-Mavzu..
Fan: Kompyuter algebrasi tizimlari O’qituvchi: T.Djiyanov
Ushbu ODEning yopiq shakldagi yechimlarini
mumkin.
topish uchun differensial Tenglamani differensial
Bu shuni ko'rsatadiki
(ko'paytuvchilarga ajratish yagona emas):

• 
  right_fact := ode_fact[2];
  

homogeneous

Berilgan differensial tenglamaning har qanday yechimi tenglamaning yechimi bo'ladi. Bu quyidagi natijaga olib keladi:

• 
  ode2 := diffop2de( right_fact, [DF,x], y(x) );
  
• 
  dsolve( ode2, y(x) );
  

operatorlarining faktorizasiyasini ishlatish operatorga o'zgartirish mumkin:

• 
  dode := de2diffop(ode,y(x), [DF,x]);
  dode := x2 DF3  ( x3  x ) DF2  2 x2 DF  x3  x
  va quyidagicha ko'paytuvchilarga ajratiladi:
  
• 
  ode_fact := DFactor( dode, [DF,x] );
  





bitta faktorlash to'g'ri ko'paytuvchiga ega bo'ladi

ode_fact :=  x DF  x ( 2  x ), DF 
2 2 2
DF x2  1 
x 
x2

right_fact := DF 
2
DF x2  1
x
x2
ode2 := 
x2
d y( x )

x 



( x2  1 ) y( x ) dx  d2
dx
2
  y( x )
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O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..
Qolgan javoblar tartibni kamaytirish orqali yuqoridagi ikkita javobdan foydalangan holda aniqlanishi mumkin. Bu aslida dsolve da avval boshdagi ODEni yechish uchun qo'llanilgan metoddir:

• 
  dsolve(ode);
  

y( x )  _C1 x BesselI( 0, x )  _C2 x BesselK( 0, x )




y( x )  _C1 x BesselI( 0, x )  _C2 x BesselK( 0, x )  _C3 x 
BesselI( 0, x ) 


 BesselK( 0, x ) e(  1/2 x2 )
x2
dx

 BesselI( 0, x ) e(  1/2 x2 )
x2

• 
  BesselK( 0, x ) 
  
  


dx 

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