Let's solve the differential equation using **power series**: \[ y' - x^2 y = 0. \] --- ## 1. **Assume a power series solution** Let \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \] Then \[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \] --- ## 2. **Substitute into the differential equation** Substitute \( y \) and \( y' \): \[ y'(x) - x^2 y(x) = 0 \] \[ \sum_{n=1}^{\infty} n a_n x^{n-1} - x^2 \sum_{n=0}^{\infty} a_n x^n = 0 \] \[ \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^{n+2} = 0 \] --- ## 3. **Re-index the second sum** Let \( m = n+2 \implies n = m-2 \): \[ \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{m=2}^{\infty} a_{m-2} x^m = 0 \] Now, write both sums in terms of \( x^{k} \): - For the first sum, let \( k = n-1 \implies n = k+1 \): \[ \sum_{k=0}^{\infty} (k+1) a_{k+1} x^{k} \] - For the second sum, \( x^m \) starts from \( m=2 \), so: \[ \sum_{k=2}^{\infty} a_{k-2} x^{k} \] So the equation becomes: \[ \sum_{k=0}^{\infty} (k+1) a_{k+1} x^k - \sum_{k=2}^{\infty} a_{k-2} x^k = 0 \] --- ## 4. **Combine into one sum** Write the first few terms explicitly to match powers: - For \( k=0 \): \( (0+1)a_1 x^0 = a_1 \) - For \( k=1 \): \( (1+1)a_2 x^1 = 2a_2 x \) - For \( k \geq 2 \): \( (k+1)a_{k+1} x^k - a_{k-2} x^k \) So, \[ a_1 + 2a_2 x + \sum_{k=2}^{\infty} \left[(k+1) a_{k+1} - a_{k-2}\right] x^k = 0 \] For this to be zero for all \( x \), each coefficient must vanish: - \( a_1 = 0 \) - \( 2 a_2 = 0 \implies a_2 = 0 \) - For \( k \geq 2 \): \((k+1) a_{k+1} - a_{k-2} = 0 \implies (k+1) a_{k+1} = a_{k-2} \) --- ## 5. **Find the recurrence relation** \[ a_{k+1} = \frac{a_{k-2}}{k+1}, \quad k \geq 2 \] Recall: - \( a_1 = 0 \) - \( a_2 = 0 \) --- ## 6. **Find the first few coefficients** Let \( a_0 \) be arbitrary (since it is not determined above). - \( a_0 \): arbitrary - \( a_1 = 0 \) - \( a_2 = 0 \) Next, \( a_3 \): \[ a_{3} = \frac{a_{0}}{3} \] \( a_4 \): \[ a_4 = \frac{a_1}{4} = 0 \] \( a_5 \): \[ a_5 = \frac{a_2}{5} = 0 \] \( a_6 \): \[ a_6 = \frac{a_3}{6} = \frac{a_0}{3 \cdot 6} = \frac{a_0}{18} \] \( a_7 = \frac{a_4}{7} = 0 \), \( a_8 = \frac{a_5}{8} = 0 \), \( a_9 = \frac{a_6}{9} = \frac{a_0}{18 \cdot 9} = \frac{a_0}{162} \), etc. So all coefficients except those with indices that are multiples of 3 are zero. Let's formalize this: --- ## 7. **General Solution** Notice: - \( a_0 \): arbitrary - \( a_3 = \frac{a_0}{3} \) - \( a_6 = \frac{a_3}{6} = \frac{a_0}{3 \cdot 6} \) - \( a_9 = \frac{a_6}{9} = \frac{a_0}{3 \cdot 6 \cdot 9} \) - In general, \[ a_{3m} = \frac{a_0}{3 \cdot 6 \cdot 9 \cdots (3m)} = \frac{a_0}{(3m)!/3^m m!} \] But let's derive it more formally: \[ a_{3m} = \frac{a_{3m-3}}{3m} \] and \( a_0 \) given. So, \[ a_{3m} = \frac{a_0}{(3m)(3m-3)\cdots 3} = \frac{a_0}{3^m m!} \] **This is because**: \[ 3m \cdot 3(m-1) \cdots 3 \cdot 1 = 3^m m! \] --- ## 8. **Final Power Series Solution** Therefore, \[ y(x) = a_0 \sum_{m=0}^\infty \frac{x^{3m}}{3^m m!} \] Or, \[ \boxed{ y(x) = a_0 \sum_{m=0}^\infty \frac{1}{m!} \left(\frac{x^3}{3}\right)^m = a_0 \exp\left(\frac{x^3}{3}\right) } \] --- ## 9. **Check:** The original equation is first order, so the general solution should have one arbitrary constant—**it does**. --- ## **Summary** **The power series solution to the differential equation \( y' - x^2 y = 0 \) is:** \[ \boxed{ y(x) = C \exp\left( \frac{x^3}{3} \right) } \] where \( C \) is an arbitrary constant. --- **This solution was derived using the power series method.**