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To solve the differential equation (x^2+4)y''+y=0 by using power series method, we can first assume that y is a power series around some point a, such as y=∑(c_n(x-a)^n) where c_n are the coefficients of the power series. We can then substitute this into the differential equation, and equate the coefficients of the powers of (x-a) to zero. This will give us a set of equations for the coefficients c_n. After substituting the power series into the differential equation and equating the coefficients of the powers of (x-a) to zero, we get: c_0(a^2+4) + c_1(2a) + c_2 = 0 c_n(a^2+4) + c_(n+1)(2a) + c_(n+2) = 0 Where n = 1, 2, 3, .... Notice that the above equation is a recurrence relation, it can be solved using characteristic equation. (a^2+4)r^2 + 2ar + 1 = 0 The roots of the characteristic equation are r = (-a± √(-a^2-4))/2 Since r is the root of the characteristic equation we know that r1.r2 = -1/4a^2 So, the general solution of the differential equation is y = c1e^(r1x) + c2e^(r2x) Where r1 and r2 are the roots of the characteristic equation. In this case, the roots of the characteristic equation are r1 = (i√3)/2, r2 = -(i√3)/2 so, the general solution of the differential equation is y = c1e^((i√3)x/2) + c2e^(-(i√3)x/2) where c1 and c2 are arbitrary constants. Hence, by using power series method we can find the general solution of the differential equation (x^2+4)y''+y=0 It is worth noting that the power series method is a technique to find the solutions of differential equations around some point (a) and it is based on the Taylor series.