1. f(t)=2+sin3t 2. f(t)=Sintcost 3. f(t)=3e2t 4. f(t)=te3t 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: Operatsion xisobning differensial tenglamalarni yechishga tadbiqi.
Misollar.
Operatsion xisobning differensial tenglamalarni yechishga tadbiqi. O’zgarmas koeffitsientli chiziqli differensial tenglama berilgan bo’lsin:
x(n)(t)+a1x(n-1)(t)+...+ an-1x(t)+ anx(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)[ antn+an-1tn-1+...+a1t+a0] = =an[tn-1x0+tn-2x0+...+x0(n-1)]+an-1[tn-2x0+tn-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.
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/t1 , t/(t2+32) cos3t
Demak
F(t)=1/91/t-1/9 t/(t2+32) 1/9 -1/9cos3t
yoki
y=1/9 -1/9cos3t.
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+Ct2_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)=21+3/2et-1/2e-t=2+3/2et-1/2e-t .
Misol. y(t)-2y(t)-3y(t)=e3t ya’ni y-2y-3y=e3t 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)=