Question

In Exercises 1-20 solve the initial value problem. Where indicated by CIG-graph the solution. 1. y" + 3y' + 2y = 6e21 + 28(1 - 1), y(0) = 2, y'(0) = -6 2. C/G Y" + y' – 2y = -106-' +58(1 - 1), y(0) = 7, y'(0) = -9 3. y" - 4y = 2e-' +58(1 - 1), y(0) = -1, y' (O) = 2 4. CIG y" + y = sin 3t + 28(t – 1/2), y(0) = 1, y'O) = -1 5. y" + 4y = 4 + 8(t - 34), y(0) = 0, y'(0) = 1 6. y" - y = 8+28(1 – 2), y(0) = -1, y'(0= 1

# 2 please

Answer 1

Given, y + y 2y = -loe + +58(t-1) ylo)= 420)=-9 Solve it by using Laplace transform > Lly") + 2ly) zly) --1011e +) 4521841 " ilg) - 4le) - L'a) - 2 kg) + 2 lo): = 100+ saº - S4e1): 21641)eos} SI S+ 25 (S2) Lly) - $(1) - (-9) + 2(3) = -10 + se-S * Cp (s: 2) 216) -75+9+)4 = -10 + se's * (S? 2) Lly) = -10 +75 - 23 + sens _(522) 116) = -10 +75 (s+1) – 23 (5+1) + se Cs?: a) Lly) ._2316.9 - 33 + 5e-5 (5+1) 5+1 y 1242 = 75-165-33 +5e-s (5+1) (S? 2) (5²2) -S 75²-165-33 + 5 es (5+1) (S? 2) (5²2) Now.. [/75²165-33.7: LS-6-13/2}+2) 13/2-6 } + 7 [S+11 (5²2) ] [ 4 (5-√2) J 1 4(5+(2) J -6-13√2. ert + 1382-6. er2t tio et - lale know that _ _ ['fe-as F (s)} = H(+-) +-aj_ where 4.lt) is heaviside step function }. HU-1). 24satasch (18[t:-)).HLE_)

- : A s. . ਦ, ੧॥ ੧-੨ - + ੧੦ 2 on ((1))
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