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Given the differential equation and its initial conditions y" + y' = 12 + 2 with y(0) = 1 and y'(0) = - 1 Use the Laplace Transform rules for derivatives to convert this function into F(S) and then solve for Y(S). L{ y(1)} = Y(s) L{ y'(t)} = S Y(s) - y (0) L{ y"(1)} = s2 Y(s) – S y(0) -y'(0)

Given the differential equation and its initial conditions y" + y' = 12 + 2 with y(0) = 1 and y'(0) = - 1 Use the Laplace Transform rules for derivatives to convert this function into F(S) and then solve for Y(S). L{ y(1)} = Y(s) L{ y'(t)} = S Y(s) - y (0) L{ y"(1)} = s2 Y(s) – S y(0) -y'(0)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Given the differential equation and its initial conditions y" + y' = t2 + 2 with y(0) = 1 and y'(0) = - 1 Use the Laplace Transform rules for derivatives to convert this function into F(s) and then solve for Y(S). L{y(1)} = Y(s) L{y'(1)} = S Y(s) - y(0) L{ y"(1)} = S² Y(s) - Sy(0) -y'(0)
Transcribed Image Text:Given the differential equation and its initial conditions y" + y' = t2 + 2 with y(0) = 1 and y'(0) = - 1 Use the Laplace Transform rules for derivatives to convert this function into F(s) and then solve for Y(S). L{y(1)} = Y(s) L{y'(1)} = S Y(s) - y(0) L{ y"(1)} = S² Y(s) - Sy(0) -y'(0)
4 - s4 A Y(s) = s4 (s + 1) 252 + S + 3 B Y(s) S3 (s2 + 1) s4 + 2s2 + 2 (C) Y(s) = s4 (s + 1) s4 s3 + 252 D Y(s) = 53 (s+ 1) (s - 1) S + 3 E Y(s) s3 (s2 + 1) s4 Y(s) = + S3 + 252 s3 (s2 + 1)
Transcribed Image Text:4 - s4 A Y(s) = s4 (s + 1) 252 + S + 3 B Y(s) S3 (s2 + 1) s4 + 2s2 + 2 (C) Y(s) = s4 (s + 1) s4 s3 + 252 D Y(s) = 53 (s+ 1) (s - 1) S + 3 E Y(s) s3 (s2 + 1) s4 Y(s) = + S3 + 252 s3 (s2 + 1)
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