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Question Number 96429 by Hassanfathi last updated on 01/Jun/20

Answered by abdomathmax last updated on 01/Jun/20

y^((3))  +y^((1))  =t+1 ⇒  L(y^((3)) )+L(y^((1)) ) =L(t+1) ⇒  t^3 L(y)−t^2 y(0)−ty^, (0)−y^((2)) (0) +tL(y)−y(0)  =L(t+1) ⇒  (t^3  +t)L(y)=t^2 y(0)+ty^′ (0)+y^((2)) (0)+y(0)+L(t+1)  L(t+1) =∫_0 ^∞  (x+1)e^(−tx)  dx  =[((x+1)/(−t))e^(−tx) ]_0 ^∞  −∫_0 ^∞ (1/(−t))e^(−tx)  dx  =(1/t) +(1/t)[−(1/t)e^(−tx) ]_0 ^∞  =(1/t) +(1/t^2 ) ⇒  (t^3 +t)L(y) =(1/t) +(1/t^2 ) +t^2 y(0)+t y^′ (0)+y^((2)) (0)+y(0)  ⇒L(y) =(1/(t(t^3  +t))) +(1/(t^2 (t^3  +t))) +(t/(t^2 +1))y(0)+(1/(t^2  +1))y^′ (0)  +((y(0)+y^((2)) (0))/(t^3  +t)) ⇒  y =L^(−1) ((1/(t(t^3  +t))))+L^(−1) ((1/(t^2 (t^3  +t))))+y(0)L^(−1) ((t/(t^2  +1)))  +y^′ (0)L^(−1) ((1/(t^2  +1)))+(y(o)+y^((2)) (0))L^(−1) ((1/(t^3 +t)))  let ddcompose f(t) =(1/(t^3  +t))  f(t) =(1/(t(t^2  +1))) =(a/t) +((bt+c)/(t^2  +1))  a =1   lim_(t→+∞)  tf(t) =0 =a+b ⇒b=−1 ⇒  f(−t)=−f(t) ⇒−(a/t) +((−bt +c)/(t^2  +1)) =−(a/t) +((−bt−c)/(t^2  +1))  ⇒c=0 ⇒f(t) =(1/t)−(t/(t^2  +1)) ⇒L^(−1) (f(t))  =1−cost also L^(−1) ((1/(t^2  +1))) =sint let decompose  g(t) =(1/(t^2 (t^3  +t))) =(1/(t^3 (t^2  +1))) =(a/t) +(b/t^2 ) +(c/t^3 ) +((αt +β)/(t^2  +1))  rest to find those coefficients....be continued...

y(3)+y(1)=t+1L(y(3))+L(y(1))=L(t+1)t3L(y)t2y(0)ty,(0)y(2)(0)+tL(y)y(0)=L(t+1)(t3+t)L(y)=t2y(0)+ty(0)+y(2)(0)+y(0)+L(t+1)L(t+1)=0(x+1)etxdx=[x+1tetx]001tetxdx=1t+1t[1tetx]0=1t+1t2(t3+t)L(y)=1t+1t2+t2y(0)+ty(0)+y(2)(0)+y(0)L(y)=1t(t3+t)+1t2(t3+t)+tt2+1y(0)+1t2+1y(0)+y(0)+y(2)(0)t3+ty=L1(1t(t3+t))+L1(1t2(t3+t))+y(0)L1(tt2+1)+y(0)L1(1t2+1)+(y(o)+y(2)(0))L1(1t3+t)letddcomposef(t)=1t3+tf(t)=1t(t2+1)=at+bt+ct2+1a=1limt+tf(t)=0=a+bb=1f(t)=f(t)at+bt+ct2+1=at+btct2+1c=0f(t)=1ttt2+1L1(f(t))=1costalsoL1(1t2+1)=sintletdecomposeg(t)=1t2(t3+t)=1t3(t2+1)=at+bt2+ct3+αt+βt2+1resttofindthosecoefficients....becontinued...

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