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

y(x)^3=exp(-x)*cos(x):

• f:=dsolve(de,y(x),series);

 
 1 y(0)2 D( y)(0)  1 x3  O( x 4 )
 2 6 
2 

 2

f : y( x)  y(0)  D( y)(0)x   1 y(0)3  1 x 2 
Eslatma: hosil bo’lgan yoyilmada D(y)(0) noldagi hosilani bildiradi: y'(0). Endi esa xususiy yechimni topish uchun boshlang’ich shartni berish qoldi:

• 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
