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Question Number 142910 by BHOOPENDRA last updated on 07/Jun/21

Answered by qaz last updated on 07/Jun/21

y(t)=t+e^(−2t) +∫_0 ^t y(τ)e^(2(t−τ)) dτ  L=(1/s^2 )+(1/(s+2))+(L/(s−2))..............L=L(y(t))(s)  ⇒L=(((s^2 +s+2)(s−2))/(s^2 (s+2)(s−3)))            =(2/3)∙(1/s^2 )+(B/s)+(4/5)∙(1/(s+2))+((14)/(45))∙(1/(s−3))  B=lim_(s→0) (d/ds)[s^2 L]=lim_(s→0) (d/ds)[(((s^2 +s+2)(s−2))/((s+2)(s−3)))]        =lim_(s→0) (([(2s+1)(s−2)+(s^2 +s+2)](s+2)(s−3)−(s^2 +s+2)(s−2)(2s−1))/([(s+2)(s−3)]^2 ))       =−(1/9)  ⇒     L=(2/3)∙(1/s^2 )−(1/9)∙(1/s)+(4/5)∙(1/(s+2))+((14)/(45))∙(1/(s−3))  ⇒      y(t)=L^(−1) =(2/3)t−(1/9)+(4/5)e^(−2t) +((14)/(45))e^(3t)

y(t)=t+e2t+0ty(τ)e2(tτ)dτL=1s2+1s+2+Ls2..............L=L(y(t))(s)L=(s2+s+2)(s2)s2(s+2)(s3)=231s2+Bs+451s+2+14451s3B=lims0dds[s2L]=lims0dds[(s2+s+2)(s2)(s+2)(s3)]=lims0[(2s+1)(s2)+(s2+s+2)](s+2)(s3)(s2+s+2)(s2)(2s1)[(s+2)(s3)]2=19L=231s2191s+451s+2+14451s3y(t)=L1=23t19+45e2t+1445e3t

Commented by BHOOPENDRA last updated on 07/Jun/21

tq sir

tqsir

Answered by Olaf_Thorendsen last updated on 07/Jun/21

  y(t) = t+e^(−2t) +e^(2t) ∫_0 ^t y(τ)e^(−2τ) dτ  y(0) = 0+e^0 +e^0 ×0 = 1  e^(−2t) y(t) = e^(−2t) t+e^(−4t) +∫_0 ^t y(τ)e^(−2τ) dτ  e^(−2t) (y′(t)−2y(t)) = e^(−2t) (1−2t)  −4e^(−4t) +y(t)e^(−2t)   y′(t)−3y(t) = 1−2t−4e^(−2t)   y(t) = (2/3)t−(1/9)+(4/5)e^(−2t) +ae^(3t)   y(0) = −(1/9)+(4/5)+a = 1 ⇒ a = ((14)/(45))  y(t) = (2/3)t−(1/9)+(4/5)e^(−2t) +((14)/(45))e^(3t)

y(t)=t+e2t+e2t0ty(τ)e2τdτy(0)=0+e0+e0×0=1e2ty(t)=e2tt+e4t+0ty(τ)e2τdτe2t(y(t)2y(t))=e2t(12t)4e4t+y(t)e2ty(t)3y(t)=12t4e2ty(t)=23t19+45e2t+ae3ty(0)=19+45+a=1a=1445y(t)=23t19+45e2t+1445e3t

Commented by BHOOPENDRA last updated on 07/Jun/21

tq sir

tqsir

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