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^-^^tsinat 9. e^-^^tcosat 10. t^n 11. tsinat 12. tcosat 13. te^-^^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+)^2+a^2] 9. (p+)/[(p+)^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+)^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
(t)X(t)-(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  (t³-t)F(t)= 3t²+2t+1-3
(t³-t)F(t)= 3t²+2t-2  F(t) = 
ni eng sodda kasrlarga ajrataylik :
= = 
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  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)=  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 

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

F(t)=  originalga o’tsak