1-mavzu. Mavzu. Differensial tenglama. Asosiy tushuncha va ta’riflar. Birinchi tartibli differensial tenglamalar uchun Koshi masalasi

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

5. f(t)=ch4t-sht 6. f(t)=2tcos3t 7. f(t)=sin²4t 8. f(t)=cos²2t

9. f(t)=2t⁵+3 10. f(t)=4t-cos2t

Quyidagi F(t) tasvir funksiyalardan f(t) original funksiyalarni toping.

11. F(t)= 12. F(t)= 13. F(t)=

14. F(t)= 15. F(t)= 16. F(t)=

17. F(t)= 18. F(t)= 19. F(t)= 20. F(t)=

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