y''-5y'-6y=e^-2t where y'(0)=1, y(0) = -1
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Question: Solve the following equation using Laplace transform.

y''-5y'-6y=e^-2t where y(0)=-1, y’(0) = 1


Laplace transforms for the 1st and 2nd differentials.
L{y’(t)} = s.Y(s) – y(0)
L{y’’(t)} = s^2.Y(s) – (s.y(0) + y’(0))


Laplace transform for the exponential.
L{e^(-2t)} = 1/(s+2)


Substituting for their Laplace transforms into the DE,
s^2.Y(s) – (s.y(0) + y’(0)) – 5(s.Y(s) – y(0)) – 6.Y(s) = 1/(s+2)


Putting in the initial condition values,
s^2.Y(s) – (-s + 1) – 5(s.Y(s) + 1) – 6.Y(s) = 1/(s+2)
(s^2 – 5s – 6)Y(s) + s – 1 – 5 = 1/(s+2)
(s – 6)(s + 1)Y(s) = 1/(s+2) – (s – 6) = (1 – (s+2)(s-6))/(s+2) = (1 – (s^2 – 4s – 12))/(s+2)
Y(s) = -(s^2 – 4s – 13)/(s+1)(s+2)(s-6)


Converting to partial fractions, this becomes
Y(s) = (1/56)*(1/(s-6)) – (8/7)*(1/(s+1)) + (1/8)*(1/(s+2))


Taking the inverse transforms,
y(t) = L^(-1){Y(s)} = (1/56)*e^(6t) – (8/7)*e^(-t) + (1/8)*e^(-2t)

Answer: y(t) = (1/56)*e^(6t) – (8/7)*e^(-t) + (1/8)*e^(-2t)

 

by Level 11 User (81.5k points)

Related questions

2 answers
1 answer
asked Sep 21, 2012 in Calculus Answers by anonymous | 292 views
0 answers
0 answers
asked Jan 17, 2012 in Calculus Answers by anonymous | 306 views
2 answers
asked Jan 4, 2012 in Calculus Answers by anonymous | 683 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,300 questions
90,708 answers
2,172 comments
99,859 users