Laplace Transformations to solve initial value problem

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Section 8.3 Solutions: In addition to the solution, I provided Y(s) so you could \"check" your work along the way. If Y(S) is not correct, you can go back and check your work up to that point before continuing to use partial fractions to expand Y (s) and taking the inverse Laplace transform of both sides. 2) 2 + 2 17(3) = L s(s2 s 6) 1 8 4 m: y = + e33 + _e21 3 15 5 4) 2 4 5 Y(S) = i (s 3)(s2 4) 1 17 2 m: y = ez' + e-Zt + -93: 4 20 5 5) 2 3 + 1 Y(S) = ; (S - 3)(32+s 2) 11 1 1 m: y = e'2' + et + _e3t 15 6 10 7) s2 + 6 . _ - HS) 2 (remember you cannot use the HeaVISIde's Method on this when (s2 + (4)02 + 1) using partial fraction decomposition to expand this function (just use regular partial fractions with no short-cut). Check out Example #4 on the webpage for Section 8.3 to review how to expand with partial fractions when there are quadratic factors in the denominator) 5 Solution: y = E sin(t) E 811101) 8.3 Exercises In Exercises 131 use the Laplace transform to solve the initial value problem. 2- y\" y' 6y = 2, 11(0) = 1, y'(0) = 0 4 y\" 43; = 2e33, y(0) = 1, y'(0) = 1 5. y\" +y' 23/ = 63$, y(0) = 1, y'(0) = 1 7 y\" + y = sin 2t, y(0) = 0, y'(0) = 1