Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
The solution to this problem is (s2+1)Y(s)−2s−1=2s2+4 .
We let L[y(t)]=Y(s) . With that, tabular Laplace-transform laws entail
L[y′]=sY(s)−y(0)=sY(s)−2
L[y′′]=s2Y(s)−sy(0)−y′(0)=s2Y(s)−2s−1
∴L[y′′+y]=(s2+1)Y(s)−2s−1 .
Since the Laplace transform of sin(kt) is ks2+k2 , we have
(s2+1)Y(s)−2s−1=2s2+4 .
This equation readily solves to Y(s)=2s3+s2+8s+6(s2+1)(s2+4) . We compute its partial fraction decomposition as Y(s)=2s+53s2+1+−23s2+4 .
Our final step is to use tabular Laplace-transform laws ( sin(kt)↦ks2+k2 , and cos(kt)↦ss2+k2 ) to get y(t)=2cost+53sint−13sin(2t) . One can check that this indeed satisfies the initial value problem.
For more information about laplace transform, visit
https://brainly.com/question/14701796
#SPJ4
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
