Menu Close

Question-96428

Tinku Tara 0 Comments
Pin




Question Number 96428 by Hassanfathi last updated on 01/Jun/20
Answered by Sourav mridha last updated on 01/Jun/20
y^(′′) (t)−3y^′ (t)+2y(t)=4 using Laplace transform s^2 y(s)−sy(0)−y^′ (0)−3sy(s)+3y(0) +2y(s)=(4/s) given y(0)=1 and y^′ (0)=0 ⇒y(s)[s^2 −3s+2]=(4/s)+s−3 ⇒y(s)=(([(4/s)+s−3])/((s−2)(s−1))) ⇒L^(−1) [y(s)]=L^(−1) [(4/(s(s−2)(s−1)))+((s−3)/((s−2)(s−1)))] ⇒y(t)=2L^(−1) [(1/s)−(((s−3))/((s−2)(s−1)))] +L^(−1) [(1/(s−1))]−L^(−1) [(1/(s−2))]+L^(−1) [(1/(s−1))] ⇒y(t)=2L^(−1) [(1/s)]−2L^(−1) [(1/((s−1)))] +2L^(−1) [(1/((s−2)))]−2L^(−1) [(1/(s−1))] +2e^t −e^(2t) ⇒y(t)=2−2e^t +e^(2t) ....(Ans)
y″(t)−3y′(t)+2y(t)=4usingLaplacetransforms2y(s)−sy(0)−y′(0)−3sy(s)+3y(0)+2y(s)=4sgiveny(0)=1andy′(0)=0⇒y(s)[s2−3s+2]=4s+s−3⇒y(s)=[4s+s−3](s−2)(s−1)⇒L−1[y(s)]=L−1[4s(s−2)(s−1)+s−3(s−2)(s−1)]⇒y(t)=2L−1[1s−(s−3)(s−2)(s−1)]+L−1[1s−1]−L−1[1s−2]+L−1[1s−1]⇒y(t)=2L−1[1s]−2L−1[1(s−1)]+2L−1[1(s−2)]−2L−1[1s−1]+2et−e2t⇒y(t)=2−2et+e2t….(Ans)
Answered by mathmax by abdo last updated on 01/Jun/20
y^(′′) (t)−3y^′ (t)+2y(t) =4 ⇒L(y^(′′) )−3L(y^′ )+2L(y) =(4/t) ⇒ t^2 L(y)−ty(0)−y^′ (0)−3(tL(y)−y(o))+2L(y) =(4/t) ⇒ (t^2 −3t +2)L(y) =(4/t) +ty(0)+y^′ (0)−3y(0) ⇒ (t^2 −3t+2)L(y) =(4/t) +t−3 ⇒L(y) =(4/(t(t^2 −3t +2))) +((t−3)/(t^2 −3t +2)) ⇒ y =4L^(−1) ((1/(t(t^2 −3t+2))))+L^(−1) (((t−3)/(t^2 −3t +2))) f(t) =((t−3)/(t^2 −3t +2)) =((t−3)/(t^2 −t−2(t−1))) =((t−3)/(t(t−1)−2(t−1))) =((t−3)/((t−1)(t−2))) =(a/(t−1)) +(b/(t−2)) ⇒a =((−2)/(−1)) =2 , b =((−1)/1) =−1 ⇒f(t) =(2/(t−1))−(1/(t−2)) ⇒ L^(−1) (f(x)) =2e^t −e^(2t) g(t) =(1/(t(t^2 −3t +2))) =(1/(t(t−1)(t−2))) =(a/t) +(b/(t−1)) +(c/(t−2)) a =(1/2) , b =−1 , c =(1/2) ⇒g(t) =(1/(2t))−(1/(t−1)) +(1/(2(t−2))) ⇒ L^(−1) (g(t)) =(1/2) −e^t +(1/2)e^(2t) ⇒y =2 −4e^t +2e^(2t) +2e^t −e^(2t) ⇒ y(t) =e^(2t) −2e^t +2
y″(t)−3y′(t)+2y(t)=4⇒L(y″)−3L(y′)+2L(y)=4t⇒t2L(y)−ty(0)−y′(0)−3(tL(y)−y(o))+2L(y)=4t⇒(t2−3t+2)L(y)=4t+ty(0)+y′(0)−3y(0)⇒(t2−3t+2)L(y)=4t+t−3⇒L(y)=4t(t2−3t+2)+t−3t2−3t+2⇒y=4L−1(1t(t2−3t+2))+L−1(t−3t2−3t+2)f(t)=t−3t2−3t+2=t−3t2−t−2(t−1)=t−3t(t−1)−2(t−1)=t−3(t−1)(t−2)=at−1+bt−2⇒a=−2−1=2,b=−11=−1⇒f(t)=2t−1−1t−2⇒L−1(f(x))=2et−e2tg(t)=1t(t2−3t+2)=1t(t−1)(t−2)=at+bt−1+ct−2a=12,b=−1,c=12⇒g(t)=12t−1t−1+12(t−2)⇒L−1(g(t))=12−et+12e2t⇒y=2−4et+2e2t+2et−e2t⇒y(t)=e2t−2et+2

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *