Problems

Solution:

1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\) then \begin{align*} y' &= \frac{\d}{\d x} \l \sum_{n=0}^\infty a_n x^n \r \\ &= \sum_{n=0}^\infty \frac{\d}{\d x} \l a_n x^n \r \\ &= \sum_{n=0}^\infty n a_n x^{n-1} \\ &= \sum_{n=1}^\infty n a_n x^{n-1} \\ \\ y'' &= \frac{\d}{\d x} \l\sum_{n=1}^\infty n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty \frac{\d}{\d x} \l n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty n(n-1) a_n x^{n-2} \\ &= \sum_{n=2}^\infty n(n-1) a_n x^{n-2} \\ \\ y &= a_0 + a_1 x+ a_2x^2 + a_3x^3 + \cdots \\ y'&= a_1 + 2a_2x+3a_3x^2 + \cdots \\ y'' &= 2a_2 + 6a_3x + \cdots \end{align*}
2. \begin{align*} && 0 &= xy''-y'+4x^3y \\ &&&= x\sum_{n=2}^\infty n(n-1) a_n x^{n-2} - \sum_{n=1}^\infty n a_n x^{n-1} + 4x^3 \sum_{n=0}^\infty a_n x^n \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=0}^\infty 4a_n x^{n+3} \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=4}^\infty 4a_{n-4} x^{n-1} \\ &&&= \sum_{n=4}^{\infty} \l n(n-1) a_n- n a_n +4a_{n-4} \r x^{n-1} + 2a_2x + 6a_3x^2-a_1-2a_2x-3a_3x^2 \\ &&&= \sum_{n=4}^{\infty} \l n(n-2) a_n +4a_{n-4} \r x^{n-1}+ 3a_3x^2-a_1 \\ \end{align*} Therefore since all coefficients are \(0\), \(a_1 = 0\), \(a_3 = 0\) and \(\displaystyle a_n = -\frac{4}{n(n-2)}a_{n-4}\). If \(a_0 = 1, a_2 = 0\), and since \(a_1 = 0, a_3 = 0\) the only values which will take non-zero value are \(a_{4k}\). We can compute these values as: \(a_{4k} = -\frac{4}{(4k)(4k-2)} a_{4k-4} = \frac{1}{2k(2k-1)}a_{4k-r}\) so \(a_{4k} = \frac{(-1)^k}{(2k)!}\), which are precisely the coefficients in the expansion \(\cos x^2\). If \(a_0 = 0, a_2 = 1\) then since \(a_1 = 0, a_3 = 0\) the only values which take non-zero values are \(a_{4k+2}\) we can compute these values as: \(a_{4k+2} = -\frac{4}{(4k+2)(4k)}a_{4k-2} = -\frac{1}{(2k+1)2k}a_{4k-2}\) so we can see that \(a_{4k+2}= \frac{(-1)^k}{(2k+1)!}\) precisely the coefficients of \(\sin x^2\)