For the given differential equation below, find the complementary function, particular integral of the function and general solution of the motion.

To find the interval of convergence for the given series, we will use the Ratio Test. The series is given by: 1 - (1)/(2)(x-3) + (1)/(3)(x-3)^2 + + (-1)^n (1)/((n+1))(x-3)^n + The general term of the series is u_n = (-1)^n (1)/(n+1)(x-3)^n. Step 1: Find the ratio | u_n+1u_n |. The (n+1)-th term is u_n+1 = (-1)^n+1 (1)/((n+1)+1)(x-3)^n+1 = (-1)^n+1 (1)/(n+2)(x-3)^n+1. | u_n+1u_n | = | (-1)^n+1 (1)/(n+2)(x-3)^n+1(-1)^n (1)/(n+1)(x-3)^n | = | (-1) · (n+1)/(n+2) · (x-3) | = (n+1)/(n+2) |x-3| Step 2: Take the limit as n . _n ( (n+1)/(n+2) |x-3| ) = |x-3| _n ( (1 + 1)/(n)1 + (2)/(n) ) = |x-3| · (1+0)/(1+0) = |x-3| Step 3: Determine the interval for convergence. For the series to converge, the limit must be less than 1: |x-3| < 1 This inequality can be written as: -1 < x-3 < 1 Add 3 to all parts of the inequality: -1+3 < x < 1+3 2 < x < 4 Step 4: Check the endpoints of the interval. At x=2: Substitute x=2 into the original series: _n=0^ (-1)^n (1)/(n+1)(2-3)^n = _n=0^ (-1)^n (1)/(n+1)(-1)^n = _n=0^ (1)/(n+1)(-1)^2n = _n=0^ (1)/(n+1) This is the harmonic series (starting from n=0, so it's 1 + (1)/(2) + (1)/(3) + ), which is known to diverge. At x=4: Substitute x=4 into the original series: _n=0^ (-1)^n (1)/(n+1)(4-3)^n = _n=0^ (-1)^n (1)/(n+1)(1)^n = _n=0^ (-1)^n (1)/(n+1) This is the alternating harmonic series. We can apply the Alternating Series Test: Let b_n = (1)/(n+1). 1. b_n > 0 for all n 0. 2. b_n is a decreasing sequence: (1)/(n+1) > (1)/(n+2). 3. _n b_n = _n (1)/(n+1) = 0. Since all conditions are met, the series converges at x=4. Step 5: State the interval of convergence. Combining the results from the ratio test and the endpoint checks, the series converges for x values such that 2 < x 4. The interval of convergence is (2, 4]. Send me the next one 📸