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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)⇒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)e−txdx=[x+1−te−tx]0∞−∫0∞1−te−txdx=1t+1t[−1te−tx]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+t⇒y=L−1(1t(t3+t))+L−1(1t2(t3+t))+y(0)L−1(tt2+1)+y′(0)L−1(1t2+1)+(y(o)+y(2)(0))L−1(1t3+t)letddcomposef(t)=1t3+tf(t)=1t(t2+1)=at+bt+ct2+1a=1limt→+∞tf(t)=0=a+b⇒b=−1⇒f(−t)=−f(t)⇒−at+−bt+ct2+1=−at+−bt−ct2+1⇒c=0⇒f(t)=1t−tt2+1⇒L−1(f(t))=1−costalsoL−1(1t2+1)=sintletdecomposeg(t)=1t2(t3+t)=1t3(t2+1)=at+bt2+ct3+αt+βt2+1resttofindthosecoefficients....becontinued...
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