Tatu samarqand filiali

Yechimning fundamental (bazisli) sistema ko’rinishi. dsolve komanda differensial tenglama yechimining fundamental sistemali (bazisli funksiyalar) ko‟rinishini topish imkonini beradi. Buning uchun dsolve komandaning parametrlarida output=basis ni ko‟rsatish kerak [17, 19].

Muammoni oydinlashtirishni mashqlarda bajarib ko‟raylik va quyidagi tadbiqlarni bajaraylik:

1. 
   misol. Quyidagi berilgan differensial tenglamaning fundamental sistema yechimini toping:
   

Yechish:
y⁽⁴⁾+2y''+y=0.

• 
  de:=diff(y(x),x$4)+2*diff(y(x),x$2)+y(x)=0;
  

de:=
y( x)
y( x)
y( x) 0

• 
  dsolve(de, y(x), output=basis);
  

cos(x),sin( x), x cos(x), x sin( x)
Koshi masalasi yoki chegaraviy masalaning yechilishi. dsolve komanda Koshi masalasi yoki chegaraviy masalaning yechimini topishi mumkin, agarda berilgan differensial tenglama uchun noaniq funksiyaning boshlang‟ich hamda chegaraviy shartlari berilsa. Boshlang‟ich yoki chegaraviy shartlarda hosilalarni ^{belgilash uchun differensial operator} D ^{ishlatiladi, masalan, y''(0)=2 shartni}

(D @@ 2)( y)(0)

2. 
   kabi berishga to‟g‟ri keladi yoki y'(1)=0 shartni:
   

D( y)(1) 0 _.

Eslatib o‟tamiz, n-chi tartibli hosila
(D@@n)( y)
kabi yoziladi.

Muammoni oydinlashtirishni mashqlarda bajarib ko‟raylik va quyidagi tadbiqlarni bajaraylik:


1-misol. Koshi masalasining yechimini toping:
y⁽⁴⁾+y''=2cosx, y(0)= 2, y'(0)=1, y''(0)=0, y'''(0)=0.
Yechish:

1. 
   de:=diff(y(x),x$4)+diff(y(x),x$2)=2*cos(x);
   

de :
y(x)
y( x)
2 cos(x)

• 
  cond:=y(0)=-2, D(y)(0)=1, (D@@2)(y)(0)=0, (D@@3)(y)(0)=0;
  

cond:=y(0)= 2, D(y)(0)=1, (D⁽²⁾)(y)(0)=0, (D⁽³⁾)(y)(0)=0

• 
  dsolve({de,cond},y(x));
  

y(x) = 2cos(x) xsin(x)+x

2. 
   misol. Quyidagi chegaraviy masalaning yechimini toping:
   

y''

Yechim grafigini yasang.

Yechish:
_, y(0)
_{0 ,} y ₂ 0 _.

• 
  restart; de:=diff(y(x),x$2)+y(x)= 2*x Pi;
  

de:=

• 
  cond:=y(0)=0,y(Pi/2)=0;
  

y(x)
y(x) 2x

cond : y(0)

• 
  dsolve({de,cond},y(x));
  

_0, y ₂ 0

y(x)=2x + cos(x)


Eslatma: yechim grafigini yasash uchun hosil bo‟lgan ifodaning o‟ng tomonini taxminan ajratishga to‟g‟ri keladi (1.2-rasm).

• 
  y1:=rhs(%): plot(y1,x=-10..20,thickness=2);
  

1.2-rasm. Chegaraviy masala yechimining grafigi.

3. 
   misol. Quyidagi oddiy differensial tenglamaning yechimini turli analitik usullar yordamida Maple dasturidan foydalanib yeching:
   

Yechish:
sin( x) y'(x)
^{cos(x) y(x) 0} .

• 
  ode_L:=sin(x)*diff(y(x),x)-cos(x)*y(x)=0;
  

ode_L := sin( x )
y( x )
cos( x ) y( x ) 0

• 
  dsolve(ode_L,[linear],useInt);
  

cos( x ) _dx
sin( x )

• 
  value(%);
  

y( x )
y( x )
_C1 e
_C1 sin( x )

• 
  dsolve(ode_L,[separable],useInt);
  

y( x )

1
dx _{_a}
d_a _C1 0

• 
  value(%);
  

ln( sin( x ) )
ln( y( x ) )
_C1 0

2. Differensial tenglamaning darajali qatorlar yordamida yaqinlashuvchi yechimlari

Ko‟pchilik differensial tenglamalar turlarining aniq analitik yechimi topilmaydi. Bu holda differensial tenglamalarning yechimini yaqinlashuvchi metodlar yordamida topish mumkin, ya‟ni noaniq funksiyani darajali qatorga yoyish orqali topish.
Differensial tenglamaning yechimini darajali qator ko‟rinishida topish uchun dsolve komandada o‟zgaruvchilardan keyin type=series (yoki shunchaki series) parametrini ko‟rsatish kerak. n-chi yoyilma tartibini ko‟rsatish uchun, ya‟ni daraja tartibini yoyilma tugaguncha, dsolve komandadan oldin tartibni aniqlaydigan Order:=n komandani qo‟yish kerak.
Agar differensial tenglamaning umumiy yechimi darajali qatorlar yoyilmasi ko‟rinishida izlanayotgan bo‟lsa, u holda topilgan yoyilmadagi x -chi daraja oldidagi koeffisiyentlar noaniq qiymatli noldagi y(0) funksiya va uning hosilalari D(y)(0), (D@@2)(y)(0) va h.k.lardan iborat bo‟ladi. Chiqarish satrida hosil bo‟lgan ifoda Makloren qatorida izlanayotgan yoyilmaga o‟xshash bo‟ladi x oldidagi koeffisiyentlar boshqacha bo‟ladi. Xususiy yechimni ajratish uchun boshlang‟ich y(0)=u1, D(y)(0)=u2, (D@@2)(y)(0)=u3 va h.k. shartlarni berishga to‟g‟ri keladi. Ushbu boshlag‟ich shartlarning soni mos differensial tenglamaning tartibiga to‟g‟ri kelishi kerak.
Darajali qatorlarga bo‟lish series tipda bo‟ladi, shuning uchun ham ushbu qator bilan yana ishlash uchun convert(%,polynom) komanda yordamida polinomga aylantirib, keyin esa hosil bo‟lgan ifodaning o‟ng tomonini rhs(%) komanda yordami bilan belgilash kerak.
Muammoni oydinlashtirishni mashqlarda bajarib ko‟raylik va quyidagi tadbiqlarni bajaraylik:
1-misol. Koshi masalasining yechimini toping:

y y xe^y _,
y(0) 0

5. 
   chi tartibli aniqlikda darajali qator ko‟rinishida yechimni izlang.
   

Yechish:

• 
  restart; Order:=5:
  
• 
  dsolve({diff(y(x),x)=y(x)+x*exp(y(x)), y(0)=0}, y(x), type=series);
  

y( x)
1 _x2
2
1 _x3
6
1 _x4
6
O( x⁵ )

Olingan natijada bo‟lganligini anglatadi.

2. 
   misol. Ushbu
   

O( x⁵ )
qo‟shiluvchi yoyilish aniqligining 5-chi tartibda


y''(x) y³(x)=ye ^xcosx

differensial tenglamaning umumiy yechimini 4-chi tartibli darajali qator yoyilmasi ko‟rinishida toping. Yoyilmani quyidagi boshlang‟ich shartlar yordamida toping:
y(0)=1, y'(0)=0.
Yechish:

• 
  restart; Order:=4: de:=diff(y(x),x$2)- y(x)^3=exp(-x)*cos(x):
  
• 
  f:=dsolve(de,y(x),series);
  

f : y( x)


y(0)


D( y)(0)x
y(0)³
1 _{x 2}
2

y(0)² D( y)(0)
1 _x3
6
O( x ⁴ )

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
6
O( x⁴ )

2. 
   misol. Koshi masalasining to 6-chi tartibli darajali qator ko‟rinishida yaqinlashuvchi yechimini hamda aniq yechimlarini toping:
   

y y 3(2
x² ) sin x _,
y(0)
¹ _, y (0)
1_, y
(0) 1_.

Aniq hamda yaqinlashuvchi yechimlarning grafigini bitta rasmda yasang.
Yechish:

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

de:=
y( x)
y(x)
3(2
x² ) sin( x)

• 
  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));
  

2
_y(x)= ²¹ cos(x)
³ x² cos(x)
2


6x sin( x) 12
7 _{e x}
4
³ _e( x)
4

• 
  y1:=rhs(%):
  
• 
  dsolve({de,cond},y(x), series);
  

y(x)= ^{1 x}
1 _x2
2
1 _x3
6
7 _x4
24
1 _x5
120
O( x⁶ )

1.2-rasm. Differensial tenglama qator ko‟rinishidagi yechimi grafigi.

Eslatma: Differensial tenglamaning qator ko‟rinishidagi yechim tipi series, shuning uchun ham keyinchalik shunday yechimni ishlatish uchun (hisoblash yoki grafikni chizish) uni albatta convert komanda yordamida polinomga konvertlash kerak (1.2- rasm).

• 
  convert(%,polynom): y2:=rhs(%):
  
• 
  p1:=plot(y1,x=-2..2,thickness=2,color=black):
  
• 
  p2:=plot(y2,x=-2..2, linestyle=3,thickness=2,
  

color=blue):

• 
  with(plots): display(p1,p2);
  

Ushbu rasmda, darajali qator yordamida yechimning eng aniq qiymati
tanlangan < x < 1 intervalda bo‟lganligi.

3.