Let's solve each of the questions **step by step**: --- ## **1. Solve the differential equation \( y'' + x^2 y = 0 \) using a power series centered at \( x = 0 \).** ### **Step 1: Assume a Power Series Solution** Let \( y(x) = \sum_{n=0}^{\infty} a_n x^n \). Then, - \( y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \) - \( y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \) ### **Step 2: Substitute into the Equation** Substitute into \( y'' + x^2y = 0 \): \[ y'' + x^2 y = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x^2 \sum_{n=0}^{\infty} a_n x^n = 0 \] \[ = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=0}^{\infty} a_n x^{n+2} = 0 \] ### **Step 3: Adjust Indices to Combine Sums** Let \( k = n-2 \) for the first sum: - When \( n = 2 \), \( k = 0 \), so \( n = k+2 \) \[ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k + \sum_{n=0}^{\infty} a_n x^{n+2} = 0 \] Now, in the second sum, let \( m = n+2 \implies n = m-2 \): \[ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k + \sum_{m=2}^{\infty} a_{m-2} x^m = 0 \] Rewriting both sums in terms of \( x^k \): \[ \sum_{k=0}^{\infty} \left[ (k+2)(k+1) a_{k+2} + a_{k-2} \right] x^k = 0 \] For \( k < 2 \), \( a_{k-2} = 0 \). ### **Step 4: Recurrence Relation** So, \[ (k+2)(k+1) a_{k+2} + a_{k-2} = 0 \implies a_{k+2} = -\frac{1}{(k+2)(k+1)} a_{k-2} \] ### **Step 5: Initial Conditions** Let \( a_0 = A \), \( a_1 = B \) (arbitrary constants). ### **Step 6: Write Out the First Few Terms** - \( a_2 = -\frac{1}{2 \cdot 1} a_{0} = -\frac{1}{2}A \) - \( a_3 = -\frac{1}{3 \cdot 2} a_{1} = -\frac{1}{6}B \) - \( a_4 = -\frac{1}{4 \cdot 3} a_{2} = -\frac{1}{12}(-\frac{1}{2}A) = \frac{1}{24}A \) - \( a_5 = -\frac{1}{5 \cdot 4} a_{3} = -\frac{1}{20}(-\frac{1}{6}B) = \frac{1}{120}B \) ### **Step 7: Write the Series Solution** \[ y(x) = A\left[1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + \cdots \right] + B\left[x - \frac{1}{6}x^3 + \frac{1}{120}x^5 + \cdots \right] \] --- ## **2. Use a power series centered at \( x = 0 \) to find the first four non-zero terms of the solution to \( y'' - 2x y' + 2y = 0 \).** **Let \( y(x) = \sum_{n=0}^{\infty} a_n x^n \).** - \( y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \) - \( y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \) Substitute into the equation: \[ y'' - 2x y' + 2y = 0 \] \[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2x \sum_{n=1}^{\infty} n a_n x^{n-1} + 2\sum_{n=0}^{\infty} a_n x^n = 0 \] \[ = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2\sum_{n=1}^{\infty} n a_n x^{n} + 2\sum_{n=0}^{\infty} a_n x^n = 0 \] Let’s shift the first sum to powers of \( x^n \): Let \( k = n-2 \Rightarrow n = k+2 \): \[ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k - 2\sum_{n=1}^{\infty} n a_n x^n + 2\sum_{n=0}^{\infty} a_n x^n = 0 \] Now, write all terms as \( x^n \), and combine: - First sum: \( (n+2)(n+1)a_{n+2}x^n \) - Second sum: \( -2n a_n x^n \) (from \( n=1 \)) - Third sum: \( 2 a_n x^n \) So, for \( n \geq 0 \): \[ (n+2)(n+1) a_{n+2} - 2n a_n + 2a_n = 0 \implies (n+2)(n+1) a_{n+2} + 2(1-n)a_n = 0 \implies a_{n+2} = -\frac{2(1-n)}{(n+2)(n+1)} a_n \] **Initial terms:** Let \( a_0 = A \), \( a_1 = B \): - For \( n=0 \): \( a_2 = -\frac{2(1-0)}{2 \cdot 1}a_0 = -\frac{2}{2}A = -A \) - For \( n=1 \): \( a_3 = -\frac{2(1-1)}{3 \cdot 2}a_1 = 0 \) - For \( n=2 \): \( a_4 = -\frac{2(1-2)}{4 \cdot 3}a_2 = -\frac{2(-1)}{12}(-A) = -\frac{2}{12}(-A) = \frac{A}{6} \) - For \( n=3 \): \( a_5 = -\frac{2(1-3)}{5 \cdot 4}a_3 = -\frac{2(-2)}{20}(0) = 0 \) - For \( n=4 \): \( a_6 = -\frac{2(1-4)}{6 \cdot 5}a_4 = -\frac{2(-3)}{30} \left(\frac{A}{6}\right) = \frac{6}{30} \cdot \frac{A}{6} = \frac{A}{30} \) ### **First Four Non-Zero Terms:** \[ y(x) = A \left[1 - x^2 + \frac{1}{6} x^4 + \frac{1}{30} x^6 + \dots \right] + Bx \] --- ## **3. Solve \( y'' + 4y = 0 \) using a power series about \( x = 0 \) and write the recurrence relation.** Assume \( y(x) = \sum_{n=0}^{\infty} a_n x^n \). - \( y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \) - \( 4y = 4\sum_{n=0}^{\infty} a_n x^n \) Write \( y'' \) as \( x^n \): Let \( k = n-2 \implies n = k+2 \): \[ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^k + 4\sum_{k=0}^{\infty} a_k x^k = 0 \] \[ \sum_{k=0}^{\infty} \left[(k+2)(k+1)a_{k+2} + 4a_k \right]x^k = 0 \] So the **recurrence relation** is: \[ a_{k+2} = -\frac{4}{(k+2)(k+1)} a_k \] --- ## **4. Classify \( x = 0 \) as an ordinary point, regular singular point, or irregular singular point for \( x^2 y'' + 3x y' + y = 0 \).** Write in standard form: \[ y'' + \frac{3}{x} y' + \frac{1}{x^2} y = 0 \] - \( P(x) = \frac{3}{x} \) - \( Q(x) = \frac{1}{x^2} \) - \( x P(x) = 3 \) (analytic at \( x=0 \)) - \( x^2 Q(x) = 1 \) (analytic at \( x=0 \)) Since both are analytic at \( x=0 \), **\( x=0 \) is a regular singular point**. --- ## **5. Determine the type of the point \( x=0 \) for \((x^2 + 1) y'' + (x-1)y' + y = 0\).** Write in standard form: \[ y'' + \frac{x-1}{x^2+1} y' + \frac{1}{x^2+1} y = 0 \] - \( P(x) = \frac{x-1}{x^2+1} \) - \( Q(x) = \frac{1}{x^2+1} \) Both \( P(x) \) and \( Q(x) \) are analytic at \( x=0 \) (denominator is 1). So, **\( x=0 \) is an ordinary point**. --- ### **Summary of Final Answers:** 1. **Power series solution:** \[ y(x) = A\left[1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + \cdots \right] + B\left[x - \frac{1}{6}x^3 + \frac{1}{120}x^5 + \cdots\right] \] 2. **First four non-zero terms:** \[ y(x) = A \left[1 - x^2 + \frac{1}{6} x^4 + \frac{1}{30} x^6 + \dots \right] + Bx \] 3. **Recurrence relation:** \( a_{k+2} = -\frac{4}{(k+2)(k+1)} a_k \) 4. **\( x=0 \) is a regular singular point.** 5. **\( x=0 \) is an ordinary point.**