I need a work check

$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$

so we have:

$$ \begin{split} y &= \sum_{n=0}^\infty C_nx^n\\ y' &= \sum_{n=1}^\infty nC_nx^{n-1}\\ y' &= \sum_{n=2}^\infty n(n-1)C_nx^{n-2} \end{split} $$

so making a substitution into the original differential equation:

$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$

$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$

$$C_{n+2} = \frac{C_n}{n+2}$$

so first a few terms:

$C_0 = C_0$ and $C_1 = C_1$ and $C_2 = \frac{C_0}{2}$ and $C_3 = \frac{C_1}{3}$ and $C_4 = \frac{C_2}{4} = \frac{C_0}{4 \cdot 2}$ and $c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$

even terms are: $\frac{C_0}{2^n \cdot n!}$ odd terms are: $\frac{C_1}{1 \cdot 3 \cdot 5 \cdot ...(2n+1)}$

and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$

Where do I go from here? if I do $y(0)$, I don't get 1. How do I take the derivative of this to check the initial condition $y'(0) = 0$

  • 1
    $\begingroup$ What do you get for $y(0)$?? You get $c_0$, right? So $c_0=1$. Same idea for $y'(0)$. Differentiate term-by-term, you did it at the beginning, do it here. $\endgroup$ – GEdgar May 10 at 21:44

You're given initial conditions so you must define $c_0$ and $c_1$ to satisfy these initial conditions.

Notice that you can write your series as

$$y(x) =c_0+ \sum_{n=1}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$

where I have simply taken the first term of the first series out front. Now it is clear, when $x=0$, both series are zero, so you are left with


so $c_0=1$. Now, take the derivative with respect to $x$ of your entire function,

$$y'(x)=\sum_{n=1}^\infty \frac{2n}{2^n \cdot n!}x^{2n-1} + \sum_{n=0}^\infty (2n+1)\frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n}$$

Now notice again when $x$ is zero, the only surviving term is the zeroeth term of the second series, giving you


and thus your entire series is

$$y = \sum_{n=0}^\infty \frac{1}{2^n \cdot n!}x^{2n}$$


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