3. Replace Laplace Transform with Elzaki Transform The Elzaki transform of a function f(t), denoted as ε[f(t)], is defined by: E[f(t)=2(p)=(t)et (3) Applying the Elzaki Transform Taking the Elzaki transform of both sides of Equation (2), we get: E [Un+1(t)] = E[un(t)] + AE | | (L(m) (un(7)) + R(un(7)) + N(un(r)) − g(7)) dr](4) Using the linearity of the Elzaki transform: Zn+1(p) = Z„(p) + XƐ [L(m) (u) + R(u„) + N(u„) − g(t)] . 4. Determining the Lagrange Multiplier A To find A, we minimize the variation of u+1(t). Taking the variation of Equation (5): λε 8(Z(p))=6(Z(p))+ɛ [z¹³ (0)] = (6) (5) This simplifies to: Solve for X: 1+AEL()(u)] = 0. (7) 1 λ= (8) EL(m) (un)] 5. Substituting Back into the Iterative Scheme Substitute A into Equation (5): 1 Zn+1(p) Zn(p)- ER(un)+N(un)-g(t)]. EL() Now, taking the inverse Elzaki transform, we get: Un+1(t) = un(t)-1 1 [ E [L (m) ( u E [R(un) + N(un) — 9(t)]] EL)(u)] 6. Elzaki Transform of the Linear Operator For the linear differential operator L (m) (u(t)) = d(), the Elzaki transform is: Thus: (9) (10) [du(t)] E p" Z(p) - pu(0) -pu' (0) - -(m-1) (0). (11) dtm EL (u)] pZ(p) - (initial conditions). Substitute this result into Equation (10) to handle the linear term. 7. Final Iterative Scheme After substituting the Elzaki transform of the linear operator, the iterative scheme for the Elzaki Variational Iteration Method (EVIM) is: 1 Zn+1(p) = Zn(p) ER(un) N(un) - g(t)]. pm Taking the inverse Elzaki transform, we get: (12) (13) Un+1(t) = u(t) - ε1 ER(un) + N(un, - N(un) — 9(t)] (14) 8. Approximate Solution The approximate solution is obtained as: u(t) = lim u,(t). (15)