1-ma’ruza: Differensial tenglamalar faniga kirish. O’zgaruv

Misol. f(t)=e^-at funksiyani tasviri topilsin. F(t) = e^-pte^-atdt == e^-(p+a)tdt=1/(p+a) (7)
Misol. f(t)=e^at funksiyani tasviri topilsin. F(t) = e^-pte^atdt == e^-(p-a)tdt=1/(p-a) (8)
Misol. f(t)=shat=1/2×(e^at-e^-at) funksiyani tasviri topilsin.
(7) va (8) dan 1/2×(e^at-e^-at) ¬¸-1/2×[1/(1-p)-1/(p+a)]=a/(p²-a²)
Demak F(t)=a/(p²-a²) yoki F(t) -¸®f(t) ya’ni shat¬¸-a/(p²-a²) (9)
Shunga o’xshash chat=(e^at+e^-at)/2 funksiya tasviri chat ¬¸- p/(p²-a²) (10)
Misol. F(t)= 7/(p²+10p+41) tasvir funksiyadan boshlangich funksiya topilsin.
Yechish: F(t)= 7/(p²+10p+41)= 7×4/[4×((p+5)²+4²)] Demak
7×4/[4×((p+5)²+4²) -¸®7/4×e^-5tsin4t. yoki F(t)-¸®7/4×e^-5tsin4t=f(t)
Misol. F(t)=(p+3)/(p²+2p+10) tasvir funksiyadan boshlang’ich
funksiya topilsin.
F(t)=(p+3)/(p²+2p+10)=[(p+1)+2]/[(p+1)²+9]=(p+1)/[(p+1)²+3²]+2/[(p+1)²+3²]=
=(p+1)/[(p+1)²+3²]+2/3×3/[(p+1)²+3²]
Demak F(t) -¸®e^-tcos3t+2/3×e^-tsin3t f(t)= e^-tcos3t+2/3×e^-tsin3t.
4. ORIGINAL-tasvir jadvali.

f(t)

F(t)= e^-ptf(t)dt

1. 1 2. sinat 3. cosat 4. cosa(t-t0) 5. e^-at 6. shat 7. chat 8. e^-^l^tsinat 9. e^-^l^tcosat 10. t^n 11. tsinat 12. tcosat 13. te^-^l^t 14. (sinat-atcosat)/2a^3 15. t^nf(t) 16. f(t)dt 17. 18. f(t-t0) 19. f¢(t) 20. f¢¢(t) 21. f¢¢¢(t) 22. f^(n)(t)

1. 1/p 2. a/(p^2+a^2) 3. p/(p^2+a^2) 4. pe^-pto/(p^2+a^2) 5. 1/p+a 6. a/(p^2-a^2) 7. p/(p^2-a^2) 8. a/[(p+l)^2+a^2] 9. (p+l)/[(p+l)^2+a^2] 10. n!/p^n+1 11. 2pa/(p^2+a^2)^2 12. -(a^2-p^2)/(p^2+a^2)^2 13. 1/(p+l)^2 14. 1/(p^2+a^2)^2 15. (-1)^nd^nF(t)/dt^n 16. F(p)/p 17. F(t)dt 18. e^-ptoF(t) 19. tF(t)-f(0) 20. t^2F(t)-[tf(0)+f¢(0)] 21. t^3F(t)-[t^2f(0)+tf ¢(0)+f¢¢(0)] 22. t^nF(t)-[t^n-1f(0)+t^n-2f¢(0)+...+ +t f^(n-2)(0)+ f^(n-1)(0)]

Quyidagi funksiyalarning F(t) tasvirni toping.

1. f(t)=2+sin3t 2. f(t)=Sint×cost 3. f(t)=3e^2t 4. f(t)=te^3t
Quyidagi F(t) tasvir funksiyalardan f(t) original funksiyalarni toping.
5. F(t)= 6. F(t)= 7. F(t)=

15-MA’RUZA:
Differensial tenglama va differensial tenglamalar sistemasini yechishning operatsion xisob usuli.
Reja:

1. 
   Operatsion xisobning differensial tenglamalarni yechishga tadbiqi.
   
2. 
   Misollar.
   

Operatsion xisobning differensial tenglamalarni yechishga tadbiqi.
O’zgarmas koeffitsientli chiziqli differensial tenglama berilgan bo’lsin:
x⁽ⁿ⁾(t)+a₁x^(n-1)(t)+...+ a_n-1x¢(t)+ a_nx(t)=f(t) (1)
bu tenglamaning x(0)=x₀, x¢(0)=x₀¢, . . . , x^(n-1)(0)=x₀^(n-1) (2)
boshlang’ich shartlarni qanoatlantiruvchi xususiy yechimni topish talab qilinsin. (1) ning har ikkala tomonidagi ifodalardan tasvirlarga o’tsak,natijada
j(t)X(t)-y(t)=F(t) (3) yoki
x(t)×[ a_ntⁿ+a_n-1t^n-1+...+a₁t+a₀] =
=a_n[t^n-1x₀+t^n-2x₀¢+...+x₀^(n-1)]+a_n-1[t^n-2x₀+t^n-3x₀¢+...+x₀^(n-2)]+. . .
. . . + a₂[tx₀+x₀¢]+a₁x₀+F(t) (3*)
tenglama hosil bo’ladi. (3) va (3*) larga yordamchi yoki tasvirlovchi yoki operator tenglama deyiladi.
(3)ni X(t) tasvirga nisbatan yechib so’ngra originalga o’tsak (1) tenglamaning (2) shartlarni qanoatlantiruvchi yechimi kelib chiqadi.
Misollar yechganda quyidagi formulalardan foydalanamiz.

x¢(t) ¬¸ tF(t)-x(0) ;
x¢¢(t) ¬¸ t²F(t)-[tx(0)+x¢(0)]
(*) x¢¢¢(t) ¬¸ t³F(t)-[t²x(0)+tx¢(0)+x¢¢(0)]
x^1V(t) ¬¸ t⁴F(t)-[t³x(0)+t²x¢(0)+tx¢¢(0)+x¢¢¢(0)]


Misol. x(0)=0 boshlang’ich shartni qanoatlantiruvchi
dx/dt+x=1 differensial tenglamani yechimi topilsin.
Yechish: xt(t)+x(t)=1
tenglamani ikki tomonini e^-tt ga ko’paytirib, [0;¥) oraliqda t bo’yicha integrallab. Laplas integraliga olib kelamiz va tasvir funktsiya F(t) ni topamiz:
e^-tt[xt(t)+x(t)]dt = e^-ttdt ,
e ^-ttxt(t)dt+ e ^-ttx(t)dt=1/t
tF(t)-x(0)+F(t)=1/t bu yerda x(0)=0.
F(t)(t+1)=1/t ; F(t)=1/[t(t+1)]=1/t-1/(t+1)
Tasvir jadvalidan F(t) -¸®1-e^-t yoki
x(t)=1-e^-t .
Misol. y(0)=yt(0)=0 boshlang’ich shartni qanoatlantiruvchi
ytt+9y=1 ( y=f(t) )
differensial tenglamani yeching.
Yechish: ytt+9y=1 tenglamani ikki tomonini e^-tt ga ko’paytirib, [0;¥) oraliqda t bo’yicha integrallab Laplas integraliga olib kelamiz va tasvir funktsiya F(t) ni topamiz:
e^-txyttdx +9 e^-txydx= e^-txdx
t²F(t)+ty(0)-yt(0)+9F(t)=1/t bu yerda y(0)=0 , yt(0)=0.
t²F(t)+9F(t)=1/t ; F(t)=1/t(t²+9)=1/9t - t/9(t²+3²)
Tasvir jadvalidan
1/t-¸®1 , t/(t²+3²) -¸® cos3t
Demak
F(t)=1/9×1/t-1/9× t/(t²+3²) -¸®1/9 -1/9×cos3t
yoki
y=1/9 -1/9×cos3t.
bu berilgan differensial tenglamani yechimi.
Misol. x¢¢¢(t)- x¢(t)=0 (4) tenglamaning
x(0)=3 ; x¢(0)=2 ; x¢¢(0)=1 (5)
boshlang’ich shartlarni qanoatlantiruvchi yechimini toping.
Avval operator tenglamasini tuzamiz. Buning uchun (4) ning chap tomonidan (*) ga ko’ra tasvirga ottamiz:
[t³F(t)-( 3t²+2t+1)]-[tF(t)-3]=0 Þ (t³-t)F(t)= 3t²+2t+1-3
(t³-t)F(t)= 3t²+2t-2 Þ F(t) = Þ
ni eng sodda kasrlarga ajrataylik :
= = Þ
3t²+2t-2 = A(t²-1)+Bt(t+1)+Ct(t^_1)
3t²+2t-2 = At²-A+Bt²+Bt+Ct^2_Ct .

t² : A+B+C=3

t : B-C=2 Þ A=2 ; B=3/2 ; C=-1/2
t⁰ : -A=-2
Shunday qilib F(t)= 2/t+3/2 ×1/(t-1) - 1/2 ×1/(t+1)
Endi jadvalga kotra originallarga o’tsak, differensial tenglamaning javobi kelib chiqadi:
x(t)=2×1+3/2×e^t-1/2×e^-t=2+3/2×e^t-1/2×e^-t .
Misol. y¢¢(t)-2y¢(t)-3y(t)=e^3t ya’ni y¢¢-2y¢-3y=e^3t differensial tenglamaning y(0)=0 ; y¢(0)=0 boshlang’ich shartni qanoatlantiruvchi xususiy yechimini toping.
Yechish. Original-tasvir jadvaliga ko’ra
t²F(t)-[ty(0)+y¢(0)]-2[tF(t)-y(0)]-3F(t)=
t²F(t)- 2tF(t) -3F(t)= F(t)(t²-2t-3)= Þ F(t)=

1=A(t+1)+B(t+1)(t-3)+C(t-3)²
1=At+A+Bt²-2tB-3B+Ct²-6Ct+9C

B+C=0 C=-B C=-B
A-2B-6C=0 A-2B+6B=0 A+4B=0
A-3B+9C=1 A-3B-9B=1 A-12B=1 Þ

Þ 16B=-1 ; B=-1/16 ; C=1/16 , A=-4B Þ A=1/4

F(t)= Þ originalga o’tsak


Foydalanilgan adabiyotlar

1. 
   Morris Tenebout, Harry Pollard. “Ordinary Differential Equations”. Birkhhauzer. Germany, 2010.
   
2. 
   Robinson J.C. “An Introduction to Ordinary Differential Equations”, Cambridge University Press 2013
   
3. 
   A.Q.O’rinov, X.N.Qosimov, Q.S.G’oziyev “Differensial tenglamalar fanidan ma’ruzalar matni” FarDU. Farg’ona-2002
   
4. 
   Филиппов А. Ф. “Сборник задач по дифференциальным уравнениям” Ижевск: Из-во РХД. 2000.
   
5. 
   Salohiddinov M.S., Nasriddinov G. “Oddiy differensial tenglamalar”.Тoshkent. O’qituvсhi. 1994.
   
6. 
   Soatov Yo.U. “Oliy matematika. Т., O’qituvсhi, 1995. 1-5 qismlar.
   
7. 
   Jabborov N.M. «Oliy matematika». 1-2 qism. Qarshi, 2010.
   
8. 
   Xurramov Sh.R. «Oliy matematika». 1-2 jild. Toshkent, “Tafakkur” nashriyoti, 2018.
   
9. 
   Yu.P.Oppoqov, B.I.Jamolov «Kompleks o’zgaruvchili funksiyalar.Operasion hisob elementlari. Matematik-fizika tenglamalaridan masalalar to’plami» Namangan 2004.
   
10. 
    Raxmonov S.K “Matematik dasturlash va optimallash usullari” fanidan Oliy ta’limning 5521900 “Informatika va axborot texnologiyalari”
    

ta’lim sohasining bakalavriat yo’nalishi talabalari uchun amaliy mashg’ulotlar uchun uslubiy qo’llanma. TATU FF. Farg’ona 2007.
11. A.Q.O’rinov, Q.S.G’oziyev, X.N.Qosimov В.44.01.00 - "FIZIKA" yo’nalishi uchun Differensial tenglamalar fanidan ma’ruzalar matni FarDU Farg’ona 2000.

Mundarija.
So’z boshi…………………………………………………………………..……3
1-ma’ruza: Differensial tenglamalar faniga kirish. O’zgaruvchilari ajralgan va ajraladigan differensial tenglamalar…………………………………………..…4
2- ma’ruza: Bir jinsli va bir jinsliga olib kelinadigan differensial tenglamalar. Amaliy masalalarga tadbiqi.(Ko’zgu masalasi)…………………………………8
3- ma’ruza: Chiziqli differensial tenglamalar va uni yechishning Lagranj va Bernulli usullari. Amaliy masalalarga tadbiqi………………………………….11
4- ma’ruza: Bernulli tenglamasi. To’la differensialli tenglamalar. Integrallovchi ko’paytuvchi……………………………………………………………………16
5- ma’ruza: Hosilaga nisbatan yechilmagan tenglamalar. Klero va Lagranj tenglamalari………………………………………………………………….…22
6- ma’ruza: Tartibini pasaytirish mumkin bo’lgan yuqori tartibli differensial tenglamalar………………………………………………………………….….26
7- ma’ruza: Yuqori tartibli chiziqli differensial tenglamalar. Vronskian. Fundamental yechim.Asosiy teoremalar…………………………………….....30
8- ma’ruza: Ikkinchi tartibli differensial tenglamalar va ularni yehishning o’zgarmasni variasiyalash usuli. Ostrogradskiy-Liuvill formulasi…………….33
9- ma’ruza: O’zgarmas koeffisientli chiziqli bir jinsli differensial tenglamalar. Xarakteristik ko’pxad…………………………………………………………..40
10- ma’ruza: O’ng tomoni maxsus ko’rinishda bo’lgan o’zgarmas koeffisientli bir jinsli bo’lmagan chiziqli differensial tenglamalar……………………….....43
11- ma’ruza: Differensial tenglamalarni analitik va taqribiy yechish usullari (matematik paketlar yordamida)…………………………………………….….47
12- ma’ruza: Differensial tenglamalar sistemasi va uni yechish usullari……..57
13- ma’ruza: Laplas almashtirishlari. Asl va tasvir……………………….…..62
14- ma’ruza: Asl va tasvirning asosiy xossalari……………………………....62
15- ma’ruza: Differensial tenglama va differensial tenglamalar sistemalarini yehichning operatsion hisob usuli……………………………………………..67
Foydalanilgan adabiyotlar……………………………………………….….70
.