10. (4 pts) In this exercise we consider finding the first five coefficients in the series solution of the first order linear initial value problem (x2 +1)y” – 6y = 0 subject to the initial condition y(0) = 3, y'(0) = 3. Since the equation has an ordinary pts at x = 0 and it has a power series solution in the form y = {cnt” no (1) Insert the formal power series into the differential equation and derive the recurrence relation Cn-2 for n=2,3,… The solution to this initial value problem can be written in the form y(x) = coy(x) +C1y2(x) where co and C are determined from the initial conditions. The function yı(x) is an even function and y2(x) is an odd function. For this example, from the initial conditions, we have co =_ and en =- n=1 The function yı (x) is an infinite series yı (x) = 1+ anx2″ NOTE note that the constant co has been fac- tored out. (2) Use the recurrence relation to find the first few coefficients of the infinite series ai=_. 22 ,43=_, 04=_NOTE note that the constant co has been factored out. Finally the polynomial y2(x) = NOTE The function y(x) is an odd degree polynomial with first term x. In other words, note that the constant c has been factored out.