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Part B (5 points each] An initial value problem y + 2y = f(©),y(0) = 0 is to be solved by Laplace transforms. (B-1) When f(t

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(B52 Apply, on lablece L{y + ay} = L{{{4}} D, we get #Ly}+9184} = F(s) & 84181-460) + 24681 = 4 (1-es) - Hes fa & (8+2) 4/3ㅇ ytay = 810 where f(t)= (0) 니냐 씨 9 지 9 F(8) = L{f(t)} 1) f(t) dt || . dt + dd See 그니 est 3 dto = ués -hes d &T IF(S) = 니US+1) - Bz ga B, (8² +28) + B2 (8+2) + 82(8+2) 8218+2) Compare coefficients of 18 and p 2 Bg = 4 a [B2=2 we get and QB 1+ Bg= A (8² +28) + Ag (8+2) + Az 82 compare Coefficients of a B- & By partial fraction - 니 A + Aq A3 + 8/8+2) S+2 8 (8+2) we get

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