Question

Solve the following IVP using the Laplace Transform. Do not leave your answer in terms of an integral. y" - y = -k8(t – 4), y(0) = 3, y'(0) = 3, kER, > 0

Answer 1

Using laplace transform we find the solution of the given differential equation.
☆ (3²1) L(Y)= sy(0) + yl(o) - Ke-45 Given the differential equation you y = -x864-4) 70-3, KER ,to Taking Laplace transform in both side Llyn)-Lly) 4(K844-u) 3 3²L(y) - sylobryl(o) -Hly)= - KL[&ltayy] [ Llynash Lly) - 59-lylo) - 52 - 1/0) -y2=60] 7 (5£j)L(*) – sy(0) -y (0)a -ke 45 L[847-ay] [ as → (32-1) L(y) = 3 5+3 ake 45 7 (5+)(5~1344) = 365+1) -ké45 3(3+1) (5+1)(5-1) 75+1) 6-1) ke

L 3 31 3+1 L() Ke 4s [ al *((y)= 3 - k . e 45 + 1. e4 5-1 I 3+1 3 e yl K 2 3-1 50 + 2 e 5+1 Taking Laplace inverse transforom we get, y(x) = 'E e 45 c10[*] - ELF THE 314673) - LI JEEP CHE 3et -45 17 3+1 - 4 bult=4) مام uft-4) et-4 [ Since, 1'1€ $F15)) = f(-a) Uffma] For, f(3) = 16 5-1 Hua 1'[H]-et L'étf(y)] = $(4-y)ult-y) =etyulty for f(b) = 14 F/H)= ('[ 3ti]et 1-'[845F(] = fltay uft-u et ryu (A-4) tay Y(A)= 3et - Kult-u) et kultay) stay This is the solution of the giver differential equation