§ XIV differensial tenglamalar va ular bilan bog’liq tushunchalar


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y1=cosx va y2=sinx funksiyalarning Vronskiy aniqlovchisi W(y1, y2) qayerda to‘g‘ri ko‘rsatilgan ?

A) W(y1, y2)=cosx+sinx . B) W(y1, y2)= cosx−sinx .

C) W(y1, y2)= cosx∙sinx . D) W(y1, y2)=1 . E) W(y1, y2)=0 .




  1. y1=ex va y2=e−x funksiyalarning Vronskiy aniqlovchisi W(y1, y2) qayerda to‘g‘ri ko‘rsatilgan ?

A) W(y1, y2)= e2x . B) W(y1, y2)= e2x . C) W(y1, y2)= −1 .

D) W(y1, y2)=0 . E) W(y1, y2)=−2 .




  1. y1=excosx va y2=exsinx funksiyalarning Vronskiy aniqlovchisi W(y1, y2) qayerda to‘g‘ri ko‘rsatilgan ?

A) W(y1, y2)= e2x(cosx+sinx) . B) W(y1, y2)= e2x(cosx+sinx) .

C) W(y1, y2)= e2xcosx∙sinx . D) W(y1, y2)= e2x . E) W(y1, y2)= e2xsin2x .




  1. y1 va y2 chiziqli bog’liq funksiyalar bo’lsa, ularning Vronskiy aniqlovchisi W(y1,y2) qaysi shartni qanoatlantiradi ?

A) W(y1,y2)=0 . B) W(y1,y2)>0 . C) W(y1,y2)<0 . D) W(y1,y2)≠0 .

E) to‘g‘ri javob keltirilmagan .




  1. Agar y1 va y2 funksiyalar differensial tenglamaning xususiy yechimlari bo‘lsa, ularining Vronskiy aniqlovchisi W(y1,y2) qaysi shartni qanoatlantirganda bu yechimlar chiziqli bog‘liq bo‘ladi ?

A) W(y1,y2)=0 . B) W(y1,y2)>0 . C) W(y1,y2)<0 . D) W(y1,y2)≠0 .

E) to‘g‘ri javob keltirilmagan .




  1. y1 va y2 funksiyalar quyidagi shartlardan qaysi birini qanoatlantirganda ularning Vronskiy aniqlovchisi W(y1,y2)=0 bo’ladi?

A) y1+y2=const. . B) y1y2=const. . C) y1y2=const. . D) y1/y2=const. .

E) keltirilgan barcha shartlarda W(y1,y2)=0 bo‘ladi .




  1. y1=sin2x va y2=1−cos2x funksiyalarning Vronskiy aniqlovchisi W(y1,y2) qayerda to’gri ifodalangan?

A) W(y1,y2)=sin2x . B) W(y1,y2)=cos2x . C) W(y1,y2)=1 .

D) W(y1,y2)=0 . E) W(y1,y2)=−1 .




  1. y1=cos2x va y2=1+cos2x funksiyalarning Vronskiy aniqlovchisi W(y1,y2) qayerda to’gri ifodalangan?

A) W(y1,y2)=sin2x . B) W(y1,y2)=cos2x . C) W(y1,y2)=0 .

D) W(y1,y2)=1 . E) W(y1,y2)=−1 .




  1. Agar y1 va y2 funksiyalar differensial tenglamaning yechimlari bo‘lsa, ularning Vronskiy aniqlochisi W(y1,y2) uchun Liuvill formulasi qanday ko‘rinishda bo‘ladi ?

A) . B) . C) .

D) . E) .




  1. Agar y1 va y2 berilgan differensial tenglamaning chiziqli erkli yechimlari, C1 va C2 ixtiyoriy o‘zgarmas sonlar bo‘lsa, bu differensial tenglmaning umumiy yechimi y qanday ko‘rinishda bo‘ladi ?

A) y= C1y1+ C2y2 . B) y= (C1+ y1)(C2+ y2) . C) y= C1y1 C2y2 .

D) y= C1y1/ C2y2 . E) y= (C1+ y1)/(C2+ y2) .




  1. bir jinsli differensial tenglamaning xususiy yechimi y qaysi ko‘rinishda izlanadi ?

A) y=xλ . B) y=sinλx . C) y=cosλx . D) y=eλx . E) y=λx .



  1. differensial tenglamaning xarakteristik tenglamasini ko‘rsating .

A) . B) . C) .

D) . E) .




  1. differensial tenglamaning y1 va y2 xususiy yechimlarining Vronskiy aniqlovchisi W(y1, y2) qaysi shartni qanoatlantirganda ular fundamental yechim bo‘lmaydi ?

A) W(y1, y2)>0 . B) W(y1, y2)<0 . C) W(y1, y2)=0 . D) W(y1, y2)≠0 .

E) ko‘rsatilgan barcha hollarda yechimlar fundamental bo‘ladi .




  1. differensial tenglamaning xarakteristik tenglamasi ildizlari λ1 , λ2 haqiqiy va λ1≠λ2 bo‘lsa, uning y1 va y2 fundamental yechimlari qayerda to‘g‘ri ko‘rsatilgan ?

A) y1=cosλ1x , y2=cosλ2x . B) y1=sinλ1x , y2=sinλ2x . C) .

D) . E) .




  1. differensial tenglamaning xarakteristik tenglamasi ildizlari λ1 , λ2 haqiqiy va λ120 bo‘lsa, uning y1 va y2 fundamental yechimlari qayerda to‘g‘ri ko‘rsatilgan ?

A) y1=cosλ0x , y2=sinλ0x . B) .

C) . D) .

E) .



  1. differensial tenglamaning xarakteristik tenglamasi ildizlari kompleks va λ1,2=α± iβ bo‘lsa, uning y1 va y2 fundamental yechimlari qayerda to‘g‘ri ko‘rsatilgan ?

A) y1=eβxcosαx , y2= eβx sinαx . B) y1=eβxcosβx , y2= eαx sinαx .

C) y1=eβxsinαx , y2= eαx cosβx . D) y1= eαx sinβx , y2= eαx cosβx .

E) y1=eαxsinαx , y2= eαx cosβx .


  1. tenglamaning umumiy yechimini toping .

A) . B) . C) .

D) . E) .




  1. , y(0)=0, y′(0)=1 Koshi masalasi yechimini toping .

A) . B) . C) .

D) ; E) .




  1. tenglamaning umumiy yechimini toping .

A) . B) . C) .

D) . E) .




  1. Koshi masalasining yecimini toping .

A) . B) . C) .

D) . E) .




  1. tenglamaning umumiy yechimini toping .

A) . B) . C) .

D) . E) .




  1. Koshi masalasining yechimini toping .

A) . B) . C) .

D) . E) .





  1. Koshi masalasi yechimining x=π/2 nuqtadagi qiymati nimaga teng ?

A) . B) . C) .

D) . E) .




§ XIV.5. II tartibli chiziqli o’zgarmas koeffitsientli bir jinslimas differensial tenglamalar



  1. II tartibli chiziqli y′′+py′+qy=f(x) differensial tenglama qaysi shartda bir jinslimas deb ataladi ?

A) f(x)=0 . B) f(x)≠0 . C) f(x)>0 . D) f(x)<0 . E) f(x)≥0 .


  1. II tartibli chiziqli y′′+py′+qy=f(x) differensial tenglama qaysi holda birjinslimas bo‘lmaydi ?

A) f(x)=0 . B) f(x)≠0 . C) f(x)>0 . D) f(x)<0 . E) f(x)≥0 .


  1. II tartibli chiziqli y′′+py′+qy=(α2−1)f(x) differensial tenglama α parametrning qanday qiymatlarida birjinslimas bo‘ladi ?

A) α >0 . B) α≠0 . C) α<0 . D) α≠±1 . E) α=±1 .


  1. II tartibli chiziqli y′′+py′+qy=(α2−1)f(x) differensial tenglama α parametrning qanday qiymatlarida birjinslimas bo‘lmaydi ?

A) α >0 . B) α≠0 . C) α<0 . D) α≠±1 . E) α=±1 .


  1. Quyidagi II tartibli chiziqli tenglamalardan qaysi biri bir jinslimas bo’ladi?

A) y′′+py′+qy=0 . B) y′′+py′+q=0 . C) y′′+py=0 .

D) y′′+qy=0 . E) keltirilgan barcha differensial tenglamalar bir jinslidir .




  1. II tartibli bir jinslimas chiziqli y′′+py′+qy=f(x) differensial tenglamaning xususiy yechimi y*, unga mos keluvchi bit jinsli tenglamaning umumiy yechimi y0 bo‘lsa, birjinslimas tenglamaninh umumiy yechimi y qanday ko‘rinishda bo‘ladi ?

A) y= y*+ y0 . B) y= y*/ y0 . C) y= y*∙y0 .

D) y=y0/ y* . E) y= C1y*+ C2 y0 .




  1. Agar II tartibli bir jinslimas chiziqli y′′+py′+qy=f(x) differensial tenglama

mos keluvchi bir jinsli tenglamaning chiziqli erkli yechimlari y1 va y2 bo‘lsa, o‘zgarmaslarni variatsiyalash usulida bir jinlimas tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=C1(x) y1 C2(x) y2 . B) y*=C1(x) y1+ C2(x) y2 . C) y*=C1(x) y1/ C2(x) y2 .

D) y*=[C1(x)+C2(x)]( y1+ y2) . E) y*=[C1(x)–C2(x)]( y1 y2) .


  1. II tartibli bir jinslimas chiziqli y′′+py′+qy=f(x) differensial tenglamaning y*

xususiy yechimini o’zgarmaslarni variatsiyalash usulida y*=C1(x)y1+C2(x)y2

ko’rinishda izlanganda (bunda y1 va y2 tegishli bir jinsli tenglamaning chiziqli erkli yechimlari) noma’lum C1(x) va C2(x) funksiyalar qaysi sistemadan topiladi?

A) . B) .

C) . D) .

E) to’g’ri javob keltirilmagan.


  1. O‘zgarmaslarni variatsiyalash usulida y′′–4y′+3y=xsin2x II tartibli chiziqli differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=C1(x)e2x+C2(x)e–2x . B) y*=C1(x)ex+C2(x)e–2x .

C) y*=C1(x)ex+C2(x)e3x . D) y*=C1(x)sin2x+C2(x)cos2x .

E) y*=C1(x)exsin2x+C2(x)e3xcos2x


  1. Agar y′′+py′+qy=Pn(x)eαx (Pn(x)–n-darajali ko‘phad) differensial tenglamada α soni λ2+pλ+q=0 xarakteristik tenglamaning ildizi bo‘lmasa va Qn(x) n-darajali ko‘phadni ifodalasa, unda differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) . B) . C) .

D) . E) .




  1. Agar y′′+py′+qy=Pn(x)eαx (Pn(x)–n-darajali ko‘phad) differensial tenglamada α soni λ2+pλ+q=0 xarakteristik tenglamaning oddiy ildizlaridan biriga teng bo‘lsa va Qn(x) n-darajali ko‘phadni ifodalasa, unda differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) . B) . C) .

D) . E) .




  1. Agar y′′+py′+qy=Pn(x)eαx (Pn(x)–n-darajali ko‘phad) differensial tenglamada α soni λ2+pλ+q=0 xarakteristik tenglamaning karrali ildizi bo‘lsa va Qn(x) biror n-darajali ko‘phadni ifodalasa, unda differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) . B) . C) .

D) . E) .


  1. y′′+5y′+4y=xex differensial tenglamaning y* xususiy yechimining ko‘rinishi qayerda to‘g‘ri ifodalangan ?

A) y*=A+Bxex . B) y*=A+Bx2ex . C) y*=(Ax+B)ex .

D) y*=x(Ax+B)ex . E) y*=x2(Ax+B)ex .




  1. y′′–4y′–5y=xe–x differensial tenglamaning y* xususiy yechimining ko‘rinishi qayerda to‘g‘ri ifodalangan ?

A) y*=A+Bxe–x . B) y*=A+Bx2e–x . C) y*=(Ax+B)e–x .

D) y*=x(Ax+B)e–x . E) y*=x2(Ax+B)e–x .




  1. y′′+py′+qy=Asinαx+Bcosαx ( soni λ2+pλ+q=0 xarakteristik tenglamaning ildizi emas) differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=sinMαx+cosMαx . B) y*=MsinαxNcosαx . C) y*=Msinαx+Ncosαx .

D) y*=x(Msinαx+Ncosαx) . E) y*=x2(Msinαx+Ncosαx) .




  1. y′′+py′+qy=Asinαx+Bcosαx (soni λ2+pλ+q=0 xarakteristik tenglamaning ildizi) differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=sinMαx+cosMαx . B) y*=MsinαxNcosαx . C) y*=Msinαx+Ncosαx .

D) y*=x(Msinαx+Ncosαx) . E) y*=x2(Msinαx+Ncosαx) .




  1. y′′+py′+qy=Asinαx ( soni λ2+pλ+q=0 xarakteristik tenglamaning ildizi emas) differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=Msinαx . B) y*=Mxsinαx . C) y*=Msinαx+Ncosαx .

D) y*=x(Msinαx+Ncosαx) . E) y*=Ncosαx .




  1. y′′+py′+qy=Bcosαx (soni λ2+pλ+q=0 xarakteristik tenglamaning ildizi) differensial tenglamaning xususiy yechimi y* qanday ko‘rinishda izlanadi ?

A) y*=Ncosαx . B) y*=Nxcosαx . C) y*=Msinαx+Ncosαx .

D) y*=x(Msinαx+Ncosαx) . E) y*=Msinαx .




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