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Use Laplace transforms to solve the initial value problem " +y 3y=0,2(0) = 2,2'(0) = 0, +z' -3z =0,4(0) = 2,%/(0) =0, where primes indicate derivatives with respect to . z(t) = y(t) = Use Laplace transforms to solve the initial value problem " +y 3y=0,2(0)=2,2'(0) = 0, +z' 3z =0,40) = 2,4(0) =0, where primes indicate derivatives with respect to , 00:40 IT LYE .Ill 34% Solution Expand =L 25- (s + 1) 2(s + 1)(s -3) (s-3)2 st - (s - 3)2 Use the linearity property of Inverse Laplace Transform: For functions f( s), g(s) and constants a, b: L {a. f(s) +b. S- (s + 1 ) = 2L 2L (s+1) (s-3) S -3) - (s-3)2 Show Steps + L S (S + 1) N/ - 2cosh V 13 t S (s - 3) 2 2 + Show Steps + 1 1 (s+1) ( S-3) - ze 2cosh 131 s* - (s -3) 2 2 sinh 13 t = 2 e 2cosh 13 t 2 -e 2 cos 11 2 2V 13 2 Show Steps + t e 2 sinh Simplify 2 e 2cosh 13 t 2 2V 13 t t V 13 t 2e 2 sinh 13 t = 2e 2 cosh 2 2 13 Leave Feedback >Show Steps + Take the partial fraction of 2(52 + 25 -6) (s + 1) ($2 + 5-3)$2 -28 - 2 s +s - 3 S L -2s - 2 +s -3 S Show Steps + S + Expand -28 - 2 -+S- 3 S+ 2 13 s + 2 13 4 S + L 2 S+ 13 13 S + 2 Use the linearity property of Inverse Laplace Transform: For functions f(s), g(s) and constants a, b: L {a. f(s) + b. S -2L 2 S - 13 2 S 13 Show Steps + S + . . e 2 cosh V13 t 2 13 2 Show Steps + L 2 V 13 t . . e sinh 13 13 2 Show Steps + L - 4t Show Steps + = 2e Zcosh ViBL) -1. e 2 2 sinh VIE + 4. + 4t Show Steps + Refine - 2e 2 cosh V13 t le 2 2 -sin V 13 t = -2e 2 cosh V 13 t 2 2 sinh V 13 t 2 e 4t 13 2 + 4