In this exercise, we consider finding the first five coefficients in the series solution of the first-order linear initial value problem (x²+2)y′′ − 6y = 0 subject to the initial condition y(0) = 1, y′(0) = 2. Since the equation has an ordinary point at x=0 and it has a power series solution in the form y = Lim(n=0→[infinity])Σ cₙxⁿ.(1) Insert the formal power series into the differential equation and derive the recurrence relation: cn= cₙ₋₂ for n=2,3,⋯ The solution to this initial value problem can be written in the form y(x)= c₀y₁(x)+c₁y₂(x) where c₀ and c₁ are determined from the initial conditions. The function y₁(x) is an even function and y₂(x) is an odd function. For this example, from the initial conditions, we have c₀ = and c₁ = . The function y₁(x) is an infinite series y₁(x) = 1+ Lim(n=1→[infinity])Σ aₙx²ⁿNote that the constant c₀ has been factored out. (2) Use the recurrence relation to find the first few coefficients of the infinite series: a₂ = , a₄ = , a₆ = , a₈ = .Note that the constant c₀ has been factored out. Finally the polynomial y₂(x)= NOTE: The function y₂(x) is an odd degree polynomial with first term x. In other words, note that the constant c₁ has been factored out.