"Ans. (3) ---2 =L ((t -1) u (t-3))=it is a type of ---L (g(t) u (t-a)) , whose solution is L (g(t+a)). (Explanation --- L (g(t) u(t-a))Let – g(t) = h(t-a) ________(1) --- L (h(t-a) u (t-a))Now, according to definition ---- L (f(t-a) u (t-a)) = L (f(t)).Hence--- L (h(t))Now, replace t by t+a in eq.(1)----g(t+a) = h(t)Hence ---L(g(t+a)))= Now, by comparison here -------2 g(t) = t – 1 a = 32 2 2 2 g(t+a) = g(t+3) = (t+3) – 1= t + 6t + 9 – 1 = t + 6t + 8(because (a+b) =2 2 a +2ab+b )2 = So, L((t – 1) u (t – 3)) = L(g(t + 3))2 2 L((t – 1) u (t – 3)) = L(t + 6t + 8)We can distribute Laplace transform on different functions.2 2 L((t – 1) u (t – 3)) = L(t ) + L(6t) + L(8) 2 2 L((t – 1) u (t – 3)) = L(t ) + 6L(t) + 8L(1)Now, by Laplace transform table -------n n+1 L[t +(s) = n!/s2 3 2 Hence ----L(t ) = 2/sL(t) = 1/sL(1) = 1/s2 3 2 So, L((t – 1) u (t – 3)) = 2/s + 6(1/s ) + 8(1/s)2 3 2 L((t – 1) u (t – 3)) = 2/s + 6/s + 8/s2 2 3 L((t – 1) u (t – 3)) = (2+6s+8s )/sAns. (21) -------Given -----y’’+y = f(t),y(0) = 0,y’(0) = -1wheref(t) = , 2, : 0 < t = 2 -{ 0, :t > 2 }Now,in general, if -----f(t) = , g(t), : 0 < t = a -{ h(t), :t > a }Then ---- f(t) = g(t) – g(t) u (t – a) + h(t) u (t – a) In this case ---f(t) = 2 – 2u(t-2)So by given conditions -----y’’ + y’ = 2 – 2u(t-2)Now, take Laplace of both side -------L *y’’ + y’+ = L *2 – 2u (t-2)] L (y’’) + L (y’) = L (2) – L (2 u (t-2)) _______ (1)Now,by definition ---2L (y’’) = s Y(s) – s y(0) – y’(0) and L (y’) = s Y(s) – y(0) And for L (2u (t – 2)), it is a type of ---L (g(t) u (t-a)) , whose solution is L (g(t+a)).So, by comparison ---- g(t) = 2anda = 2 L (2u (t – 2)) = L (g(t+2)) = L (2)Now, by equation (1) -----------------2 s Y(s) – s y(0) – y’(0) + s Y(s) – y(0)= L (2) – L (2) 2 s Y(s) – s (0) – (-1)+ s Y(s) – (0) = 02 s Y(s) + 1 + s Y(s) = 02 (s + s) Y(s) = -12 Y(s) = -1/(s + s)Now, by taking inverse of Laplace -------1 -1 2 L [Y(s)] = L [-1/(s + s)]-1 -1 L [Y(s)] = L [-1/s(s+1)]-1 -1 L [Y(s)] = L [(s – (s+1))/s(s+1)]-1 -1 L [Y(s)] = L [{1/(s+1)} – {1/s}]-1 -1 -1 L [Y(s)] = L (1/(s+1)) – L (1/s) "