We are asked to find a series solution about (x = 1) for [ x(1-x)\frac{d^2y}{dx^2} - (1+3x)\frac{dy}{dx} - y = 0. ] Since (x = 1) is a regular point, we can attempt a Frobenius-type series centered at (x = 1). Let [ t = x - 1, \quad y = \sum_{n=0}^{\infty} a_n t^n. ] Then (\frac{dy}{dx} = \sum_{n=1}^{\infty} n a_n t^{n-1}) and (\frac{d^2y}{dx^2} = \sum_{n=2}^{\infty} n(n-1) a_n t^{n-2}). Rewrite the differential equation in terms of (t): [ x = t+1 \implies x(1-x) = (t+1)(1-(t+1)) = (t+1)(-t) = -t(t+1) = -t^2 - t. ] Also, (-(1+3x) = -(1 + 3(t+1)) = -(4 + 3t) = -4 - 3t). So the equation becomes [ (-t^2 - t) \frac{d^2y}{dt^2} + (-4 - 3t)\frac{dy}{dt} - y = 0 \implies (t^2 + t) y'' + (4 + 3t) y' + y = 0. ] Substitute the series: 1. ( (t^2 + t) y'' = \sum_{n=2}^{\infty} n(n-1) a_n t^{n-2} (t^2 + t) = \sum_{n=2}^{\infty} n(n-1) a_n (t^n + t^{n-1}) ) (\implies \sum_{n=2}^{\infty} n(n-1)a_n t^n + \sum_{n=2}^{\infty} n(n-1) a_n t^{n-1}) 2. ( (4 + 3t) y' = 4 \sum_{n=1}^{\infty} n a_n t^{n-1} + 3 \sum_{n=1}^{\infty} n a_n t^n = \sum_{n=1}^{\infty} 4 n a_n t^{n-1} + \sum_{n=1}^{\infty} 3 n a_n t^n ) 3. ( y = \sum_{n=0}^{\infty} a_n t^n ) Combine coefficients of (t^k) for recurrence relation: * For (t^0) (n = 0 term): Only terms from (\sum_{n=2}^{\infty} n(n-1)a_n t^{n-1}) with n=2 gives (2a_2 t^1) (so not t^0), (\sum 4 n a_n t^{n-1}) with n=1 gives 4a_1, (y = a_0). So (t^0) coefficient: (4a_1 + a_0 = 0 \implies a_1 = -\frac{a_0}{4}). * For (t^1) (n = 1): (\sum n(n-1)a_n t^{n-1} ) with n=2 gives (2a_2 t^1) (\sum 4 n a_n t^{n-1}) with n=2 gives (8 a_2 t^1) (\sum n(n-1) a_n t^n) with n=1 gives 0 (\sum 3 n a_n