Question

x" +0.4x' + 2x = 1 - Hz(t)
solve using laplace transforms

Answer 1

CA V" +0.4x' +2x = 1 - Hş (+) Taking Laplace Transformation Worit. t. ["] +0.4 d [x"] +2 d [x] = { [t] - d [H5 (4)] det, [x(b)] = X). Therefore, (s**@)- 546)-7'%) 40.4 (s®() – 26) +2XC) - 5 - éss +0.45 +2) X6) = 4-50% [ * *6=07 (4+*+2)x9 = 14 X09= 5(8) Taning inveree haplace Transformatie 12 +0. 4s +2 ° x (0)= 01 L x'(o)=0 -SS. + 2 WoW 8" (10): f'(**) -- ( ST 5

7 x (t) = f(t) – H5 Ct) f The f6)-1(10:] 2 Rc1 2 25 5 +2 3) 1 = A (3+ x + 2) + (B5+C)s. Comparing the co-efficients of 1,5, 52 1: 1 = 2A = A= s: 0= 2^ + C = C = 2 - - Š: 0 = A + B = B = -A = - IN

2 Ś 10 • 3-4 (4746"[***--*-043-45). ferte e o 494) - to ņē simu 569 = 4- e *((*) - tip mo siya ng The Solution of the ODE , * (t) = f() – Hy (6) f(-5) Pathore, f()=4- Le Mans pe sine 4/5 xls Where sin
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