Let's solve the differential equation using **power series**: \[ y'' - x y' - 2x y = 0 \] ### **Step 1: Assume a Power Series Solution** Assume: \[ y(x) = \sum_{n=0}^\infty a_n x^n \] Calculate derivatives: \[ y'(x) = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n \] \[ y''(x) = \sum_{n=2}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n \] ### **Step 2: Substitute into the Equation** \[ y'' - x y' - 2x y = 0 \] Substitute the series: - \(y'' = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n\) - \(x y' = x \sum_{n=0}^\infty (n+1)a_{n+1} x^n = \sum_{n=0}^\infty (n+1)a_{n+1} x^{n+1} = \sum_{n=1}^\infty n a_n x^n\) - \(2x y = 2x \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty 2a_n x^{n+1} = \sum_{n=1}^\infty 2a_{n-1} x^n\) So, the equation becomes: \[ \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^n - \sum_{n=1}^\infty n a_n x^n - \sum_{n=1}^\infty 2a_{n-1} x^n = 0 \] ### **Step 3: Combine into One Series** Write all sums in terms of \(x^n\): \[ (n+2)(n+1)a_{n+2} - n a_n - 2a_{n-1} = 0 \quad \text{for } n \geq 1 \] For \(n = 0\): \[ (n+2)(n+1)a_{n+2}\Big|_{n=0} - 0 - 0 = 2a_2 = 0 \implies a_2 = 0 \] For \(n \geq 1\): \[ (n+2)(n+1)a_{n+2} = n a_n + 2a_{n-1} \] \[ a_{n+2} = \frac{n a_n + 2a_{n-1}}{(n+2)(n+1)} \] ### **Step 4: Find the Coefficients** Let’s start with \(a_0\) and \(a_1\) as arbitrary constants: #### \(n = 0:\) \(a_2 = 0\) #### \(n = 1:\) \[ a_3 = \frac{1 a_1 + 2 a_0}{3 \cdot 2} = \frac{a_1 + 2a_0}{6} \] #### \(n = 2:\) \[ a_4 = \frac{2 a_2 + 2 a_1}{4 \cdot 3} = \frac{2 a_1}{12} = \frac{a_1}{6} \] (because \(a_2 = 0\)) #### \(n = 3:\) \[ a_5 = \frac{3 a_3 + 2 a_2}{5 \cdot 4} = \frac{3 a_3}{20} \] (because \(a_2 = 0\)) Substitute \(a_3\): \[ a_5 = \frac{3}{20} \cdot \frac{a_1 + 2a_0}{6} = \frac{a_1 + 2a_0}{40} \] #### \(n = 4:\) \[ a_6 = \frac{4 a_4 + 2 a_3}{6 \cdot 5} = \frac{4 a_4 + 2 a_3}{30} \] Substitute \(a_4\) and \(a_3\): \[ a_6 = \frac{4 \cdot \frac{a_1}{6} + 2 \cdot \frac{a_1 + 2a_0}{6}}{30} = \frac{\frac{4a_1 + 2a_1 + 4a_0}{6}}{30} = \frac{\frac{6a_1 + 4a_0}{6}}{30} = \frac{a_1 + \frac{2}{3}a_0}{30} \] ### **Step 5: Write the Solution** So, the power series solution is: \[ \boxed{ y(x) = a_0 + a_1 x + 0 \cdot x^2 + \frac{a_1 + 2a_0}{6} x^3 + \frac{a_1}{6} x^4 + \frac{a_1 + 2a_0}{40} x^5 + \left(\frac{a_1 + \frac{2}{3}a_0}{30}\right) x^6 + \cdots } \] Where \(a_0\) and \(a_1\) are arbitrary constants determined by initial conditions. --- **Summary:** **The power series solution to \(y'' - x y' - 2x y = 0\) is** \[ \boxed{ y(x) = a_0 + a_1 x + \sum_{n=2}^\infty a_n x^n } \] **with the recurrence** \[ a_{n+2} = \frac{n a_n + 2 a_{n-1}}{(n+2)(n+1)} \] **and \(a_2 = 0\), \(a_3 = \frac{a_1 + 2a_0}{6}\), etc.**

A power series solution involves assuming the solution can be expressed as an infinite sum of terms \( y(x) = \sum_{n=0}^\infty a_n x^n \), substituting into the differential equation, and deriving a recurrence relation for the coefficients. This method transforms the differential equation into an algebraic recurrence, allowing the construction of the solution as a series expansion around a point.