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dsolve('eqn1', 'eqn2',...)- |
| bet | 107/141 | | Sana | 15.01.2024 | | Hajmi | | | #138013 |
Bog'liq KM majmua (1)dsolve('eqn1', 'eqn2',...)-boshlang'ich shakllarga ega bo'lgan differensial tenglamalar
sistemasining analitik yechimini qaytaradi.Avval tenglamalar keyin boshlang'ich
shakllar eqni tengliklr kurinishida beriladi.Tenglik belgilari quyilmagan ifodalar
nolga teng, deb olinadi. Sukut bo'yicha ekran (mustaqil) o'zgaruvchi sifatida odatda
vaqtni ifodolovchi t o'zgaruvchi olinadi.Agar erkin o'zgaruvchi sifatida boshqa
o'zgaruvchi olinsa y dsolve funksiyasi parametrlari ruyxatining oxiriga qushib
quyiladi.Ifodalarda D simvolli bilan erkin o'zgaruvchi buyicha xosila belgilanadi,
ya'ni d/dt, D2 ESA
ni bildiradi va h.k.Erkin uzgaruvchilarning nimi D bilan
boshlanmasligi kerak.
Boshlang'ich shartlar 'y(a)=b' yoki 'Dy(a)=b' tengliklar kurinishida beriladi, bu
yerda y - bog'liq uzgaruvchi, a va b - konstantalar ular simvolli ham bulishi
mumkin.Tenglamalardagi
konstantalar
ham
simvolli
bulishi
mumkin.Agar
boshlang'ich shartlar soni differensial tenglamalar sonidan kam bulsa, yechimda C1,
C2 va h.k.Ixtiyoriy doimiylar mavjud bo'ladi.
dsolve funksiyasidan foydalanishga misollar.
1-misol
x"=-2x'
differensial tenglamani yechish
>>dsolve(‘D2x=-2*x')
ans=
C1*cos(2^(1/2)*t)+C2*sin(2^(1/2)*t)
yoki
C1cos
2-misol
y’’=-ax+y’, y(0)=b
differensial tenglamani yechish
>>dsolve('D2y=-a*x+y','(0)=b','x')
ans=
a*x+C1*sinh(x)+b*cosh(x)
yoki
ax+C1sinh(x)+b cosh(x)
257
3-misol
differensial tenglamani yechish va yechimni tekshirish
>>syms x
>>S=dsolve('D4y-y-5*exp(x)*sin(x)-x^4','x')
s =
149/208*cos(x)*exp(x)-24-x^4-57/104*exp(x)*sin(x)-21/26*exp(x)*sin(x)*sos(x)^2-
1/4*sin(x)*exp(x)*sin(s*x)+1/2*sin(x)*exp(x)*cos(2*x)-
41/52*cos(x)^3exp(x)+15/208*cos(3*x)*exp(x)-
5/104*sin(3*x)*exp(x)+C1*exp(x)+C2*sin(x)+C3*cos(x)+C4*exp(-x)
>>[R,HOW]=simple(S)
R=
-24-x^4-exp(x)*sin(x)+C1*exp(x)C2*sin(x)+C3*cos(x)+C4*exp(-x)
yechimni tekshirish:
>>diff(R,x,4)-R-5*exp(x)*sin(x)-x^4
ans=
0
>>syms x
>> S=dsolve('D3y+2*D2y+Dy=-2*exp(-2*x)','y(0)=2','Dy(0)=1','D2y(0)=1','x')
S =
exp(-2*x)+4-3*exp(-x)
yechimni tekshirish
>>diff(S,x,3)+2*diff(S,x,2)+diff(S,x)
ans=
-2*exp(-2*x)
Boshlang'ich shartlarning bajarilishini tekshirish
>>subs(s,x,o)
ans=
2
>>subs(diff(S,x),x,0)
ans=
1
>>subs(diff(S,x,2),x,0)
asn =
1
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