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

Differensial tenglamalar sistemasini echish
Quyidagi misollarda ikkita differensial tenglamalar sistemasi ikki xil usulda echilgan. YA’ni qatopga yoyish yoli bilan va Laplas almashtirishlaridan foydalanish yoli bilan. Shuni ta’kidlash lozimki, differensial tenglama qatorga yoyish yoli bilan echilganda taqribiy yechim topiladi. SHuning uchun ikki usulda topilgan echimlar farq qilmoqda.
y( x )  
1
 k  
 m 
 k x 
 
e  m 
1  e  1  e 
 
 k 
 m 
Fan: Kompyuter algebrasi tizimlari
O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..

• 
  sys:=diff(y(x),x)=2*z(x)-y(x)-x,diff(z(x),x)=y(x);
  
• 
  fens:={y(x),z(x)};
  
• 
  dsolve({sys,y(0)=0,z(0)=1},fens);
  
• 
  dsolve({sys,y(0)=0,z(0)=1},fens,series);
  

sys :=  y( x )  2 z( x )  y( x )  x,  z( x )  y( x )
x x
fens := { z( x ), y( x )}
{ z( x )  5 e ( 2 x )  1 e x  1  1 x, y( x )   5 e ( 2 x )  1 e x  1 } 12 3 4 2 6 3 2
{ y( x ) 
2 x  3 x2  7 x3  13 x4  9 x5  53 x6  107 x7 
x8 
71 61
13440 51840
x9  O( x10 ),
2 6 24 40 720 5040
z( x ) 
53 107 71
1  x2  1 x3  7 x4  13 x5  3 x6  x7  x8  x9  O( x10 ) }
2 24 120 80 5040 40320 120960

• 
  dsolve({sys,y(0)=0,z(0)=1},fens,laplace);
  { z( x )  1 e x  5 e ( 2 x )  1 x  1, y( x )   5 e ( 2 x )  1 e x  1 } 3 12 2 4 6 3 2
  Maplening differensial tenglamalarni echish bo’yicha juda katta imkoniyatlari bo’lsa ham bu degan so’z har qanday differensial tenglamani analitik usulda echa oladi degani emas. Bunday hollarda conli usulda echish mumkin bo’ladi.
  

Fan: Kompyuter algebrasi tizimlari

O’qituvchi: T.Djiyanov
II-kurs
7-Mavzu..