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Hw13-3.2-LT-IVP: Problem 3 Problem Value: 10 point(s). Problem Score: 43%. Attempts Remaining: 20 attempts. Help Entering Answers See Example 3.2.2, in Section 3.2, in the MTH 235 Lecture Notes. (10 points) Consider the initial value problem for function y given by y" - 6y + 10 y = 0, y(0) = 2, 7(0) = 5. Part 1: Finding Y() - Part 2: Rewriting Y(s) (b) The characteristic polynomial in the denominator of Y(8) above has complex roots. Therefore, complete the square in that denominator and rewrite Y(s) as follows, A(s - a) + B Y(s) =- ((s - 2)2 + B2) where A, B, a, B are constants, with p > 0. Find the the constants A, B, a, B. A = 2 2 B = 19 a = 3 2 = sqrt(19) Part 3: Finding y(t) Help Entering Answers Preview My Answers Submit Answers Show me another

Answer 1

Given yo - +o0y = 0, yomd=a, ykm5 By Laplace transformation {y"(t)}= s²res) - syco) y'ro) 14 s')} = S YS)- 9 to) l{408)} = Yes - y" - byl + loy - By applying caplale on broso 5 SY15)-SY) - YO) - Slsy(s) -(0)] +10Y68) - (-65+10]YCS) – 5(2) - 5 +612) = 0 Y(s) 25–7 25-7 S2-65+10 5-65 +9 -9410 25 -7 218-3+3)-? (5-3)2+1 (5-33241 - 215-3) +6- (5-3)² +1 Yes) • 205-3) -1 (5-332 + 1² f compare with Y(S) = Alsod) + B TS-48² 482 Thus 1 A=2 B= -1