Let's tackle the problem step by step for the differential equation: \[ 2x y'' + y' - x y = 0 \] --- ## (a) **Show that \( x_0 = 0 \) is a regular singular point.** A second-order linear ODE: \[ y'' + P(x)y' + Q(x)y = 0 \] has a **regular singular point at \( x_0 \)** if \( (x-x_0)P(x) \) and \( (x-x_0)^2 Q(x) \) are analytic at \( x_0 \). Let's write our equation in standard form: \[ 2x y'' + y' - x y = 0 \implies y'' + \frac{1}{2x} y' - \frac{1}{2} y = 0 \] So, - \( P(x) = \frac{1}{2x} \) - \( Q(x) = -\frac{1}{2} \) Calculate: - \( (x-0)P(x) = x \cdot \frac{1}{2x} = \frac{1}{2} \) (analytic at \( x=0 \)) - \( (x-0)^2 Q(x) = x^2 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} x^2 \) (analytic at \( x=0 \)) **Conclusion:** Both are analytic at \( x=0 \), so \( x=0 \) is a **regular singular point**. --- ## (b) **Determine the roots \( r_1 \) and \( r_2 \) of the indicial equation.** For a regular singular point at \( x=0 \), the indicial equation is: \[ r(r-1) + p_0 r + q_0 = 0 \] where: \[ p_0 = \lim_{x\to 0} x P(x) = \frac{1}{2} \] \[ q_0 = \lim_{x\to 0} x^2 Q(x) = 0 \] So the indicial equation becomes: \[ r(r-1) + \frac{1}{2} r = 0 \implies r^2 - r + \frac{1}{2} r = 0 \implies r^2 - \frac{1}{2} r = 0 \implies r(r - \frac{1}{2}) = 0 \] So the roots are: \[ \boxed{r_1 = 0, \quad r_2 = \frac{1}{2}} \] --- ## (c) **Find the series solutions \( y_1 \) and \( y_2 \) corresponding to \( r_1 \) and \( r_2 \) (first four nonzero terms of each)** We look for solutions of the form: \[ y(x) = x^r \sum_{n=0}^{\infty} a_n x^n \] Let’s plug into the ODE. ### **Step 1: Write derivatives** \[ y(x) = x^r \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+r} \] \[ y'(x) = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} \] \[ y''(x) = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} \] ### **Step 2: Substitute into the equation** Recall: \[ 2x y'' + y' - x y = 0 \] Plugging in: - \( 2x y'' = 2x \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} = 2 \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-1} \) - \( y' = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} \) - \( -x y = -x \sum_{n=0}^{\infty} a_n x^{n+r} = -\sum_{n=0}^{\infty} a_n x^{n+r+1} \) Sum: \[ 2 \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-1} + \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} - \sum_{n=0}^{\infty} a_n x^{n+r+1} = 0 \] Combine first two terms: \[ \sum_{n=0}^{\infty} a_n \big[ 2(n+r)(n+r-1) + (n+r) \big] x^{n+r-1} - \sum_{n=0}^{\infty} a_n x^{n+r+1} = 0 \] \[ \sum_{n=0}^{\infty} a_n \big[2(n+r)^2 - 2(n+r) + (n+r) \big] x^{n+r-1} - \sum_{n=0}^{\infty} a_n x^{n+r+1} = 0 \] \[ \sum_{n=0}^{\infty} a_n \big[2(n+r)^2 - (n+r) \big] x^{n+r-1} - \sum_{n=0}^{\infty} a_n x^{n+r+1} = 0 \] Let’s shift the second sum to match the powers: Let \( k = n+2 \Rightarrow n = k-2 \): \[ \sum_{n=0}^{\infty} a_n \big[2(n+r)^2 - (n+r) \big] x^{n+r-1} - \sum_{k=2}^{\infty} a_{k-2} x^{k+r-1} = 0 \] Now, write both sums in terms of \( x^{s} \) with \( s = n+r-1 \): First sum: \( n \) from \( 0 \) to \( \infty \) Second sum: \( k = n+r-1 \) as in first sum, but \( k \geq 2 \implies n \geq 2 \) So, for \( n = 0, 1 \), only the first sum contributes. For \( n \geq 2 \): \[ a_n \big[2(n+r)^2 - (n+r) \big] - a_{n-2} = 0 \] So, \[ a_n = \frac{a_{n-2}}{2(n+r)^2 - (n+r)} \] --- ### **Step 3: Find solutions for each root** #### **Case 1: \( r_1 = 0 \)** \[ a_n = \frac{a_{n-2}}{2 n^2 - n} \] Set \( a_0 \) arbitrary (let \( a_0 = 1 \) for simplicity), \( a_1 \) arbitrary (let's choose \( a_1 = c \), but let's first check the recurrence for \( n=1 \)). **For \( n=1 \):** \[ a_1 \text{ is arbitrary} \] **For \( n=2 \):** \[ a_2 = \frac{a_0}{2 \cdot 2^2 - 2} = \frac{a_0}{8 - 2} = \frac{a_0}{6} \] **For \( n=3 \):** \[ a_3 = \frac{a_1}{2 \cdot 3^2 - 3} = \frac{a_1}{18 - 3} = \frac{a_1}{15} \] **For \( n=4 \):** \[ a_4 = \frac{a_2}{2 \cdot 4^2 - 4} = \frac{a_2}{32 - 4} = \frac{a_2}{28} \] But \( a_2 = a_0/6 \): \[ a_4 = \frac{a_0/6}{28} = \frac{a_0}{168} \] **For \( n=5 \):** \[ a_5 = \frac{a_3}{2\cdot 5^2 - 5} = \frac{a_3}{50-5} = \frac{a_3}{45} \] But \( a_3 = a_1/15 \): \[ a_5 = \frac{a_1/15}{

Let's carefully work through the problem step by step. --- ## Given differential equation: \[ 2x y'' + y' - x y = 0 \] --- ## (a) Show that \(x_0 = 0\) is a regular singular point. **Step 1: Write the equation in standard form:** \[ y'' + P(x) y' + Q(x) y = 0 \] Divide through by \(2x\): \[ y'' + \frac{1}{2x} y' - \frac{1}{2} y = 0 \] So: \[ P(x) = \frac{1}{2x}, \quad Q(x) = -\frac{1}{2} \] **Step 2: Check the conditions for regular singularity at \(x=0\):** \[ (x - 0) P(x) = x \cdot \frac{1}{2x} = \frac{1}{2} \quad \text{(analytic at \(x=0\))} \] \[ (x - 0)^2 Q(x) = x^2 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} x^2 \quad \text{(analytic at \(x=0\))} \] Since both are analytic at \(x=0\), **\(x=0\) is a regular singular point**. --- ## (b) Find the roots \(r_1\) and \(r_2\) of the indicial equation. **Step 1: Form the indicial equation:** For a regular singular point, the indicial equation is: \[ r(r-1) + p_0 r + q_0 = 0 \] where: \[ p_0 = \lim_{x \to 0} x P(x) = \lim_{x \to 0} x \cdot \frac{1}{2x} = \frac{1}{2} \] \[ q_0 = \lim_{x \to 0} x^2 Q(x) = x^2 \cdot \left(-\frac{1}{2}\right) = 0 \] **Step 2: Write the indicial equation:** \[ r(r-1) + \frac{1}{2} r = 0 \] \[ r^2 - r + \frac{1}{2} r = 0 \] \[ r^2 - \frac{1}{2} r = 0 \] \[ r(r - \frac{1}{2}) = 0 \] **Solutions:** \[ \boxed{ r_1 = 0, \quad r_2 = \frac{1}{2} } \] --- ## (c) Find the series solutions \(y_1\) and \(y_2\) corresponding to \(r_1\) and \(r_2\). Assume solutions of the form: \[ y(x) = x^{r} \sum_{n=0}^{\infty} a_n x^{n} \] --- ### **Case 1: \( r = 0 \)** **Step 1: Express derivatives:** \[ y = \sum_{n=0}^\infty a_n x^{n} \] \[ y' = \sum_{n=1}^\infty n a_n x^{n-1} \] \[ y'' = \sum_{n=2}^\infty n(n-1) a_n x^{n-2} \] **Step 2: Substitute into the original ODE:** \[ 2x y'' + y' - x y = 0 \] becomes: \[ 2x \sum_{n=2}^\infty n(n-1) a_n x^{n-2} + \sum_{n=1}^\infty n a_n x^{n-1} - x \sum_{n=0}^\infty a_n x^{n} = 0 \] Simplify each term: - \(2x y'' = 2 \sum_{n=2}^\infty n(n-1) a_n x^{n-1}\) - \(y' = \sum_{n=1}^\infty n a_n x^{n-1}\) - \(-x y = - \sum_{n=0}^\infty a_n x^{n+1}\) --- ### **Step 3: Write all sums with the same power of \(x\):** Express the last