Please use typed out written symbol form and show work if possible. Thanks! Use
ID: 1946878 • Letter: P
Question
Please use typed out written symbol form and show work if possible. Thanks!
Use the Laplace transform method to solve the differential equation: y"-y' + 6y = e-t with initial condition: y(0) = 0; y'(0) = 2Explanation / Answer
y''-y'+6y= exp(-t) , y(0)=0, y'(0)=2 L{ y''-y'+6y= exp(-t) } s^2 L{y} -s y(0) - y'(0) - (s L{y} -y(0) ) + 6 L{y} = 1/(s+1) s^2 L{y} - 2 - s L{y} + 6 L{y} = 1/(s+1) L{y} (s^2 - s +6) = 2 + 1/(s+1) L{y} ((s-3)(s+2)) = 2 + 1/(s+1) L{y} = 2 / ((s-3)(s+2)) + 1/ ( (s-3)(s+2) (s+1) ) y(t) = L^(-1) { 2 / ((s-3)(s+2)) + 1/ ( (s-3)(s+2) (s+1) ) } 2 / ((s-3)(s+2)) = A / (s-3) + B/(s+2) A= -B==> A=2/5 , B= /2/5 L^(-1) { 2 / ((s-3)(s+2)) } = L^(-1) { A / (s-3) + B/(s+2) } = (2/5) exp(3t) - (2/5) exp(-2t) 1/ ( (s-3)(s+2) (s+1) ) = A / (s-3) + B/(s+2) + C/(s+1) A= (1/10) , B= (2/5) , C= -1/2 L^(-1) { A / (s-3) + B/(s+2) + C/(s+1) } = (1/10) exp(3t) + (2/5) exp(-2t) -(1/2) exp(-t) y(t) = L^(-1) { 2 / ((s-3)(s+2)) + 1/ ( (s-3)(s+2) (s+1) ) } = (2/5) exp(3t) - (2/5) exp(-2t) + (1/10) exp(3t) + (2/5) exp(-2t) -(1/2) exp(-t) y(t) =(2/5) exp(3t) - ( 1/2) exp(-t) ANSWER