# Solving differential equation using power series $y'' - xy' -y = 0$ given $y(0) = 1$ and $y'(0) = 0$

I think I made a mistake somewhere:

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

so we have:

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

so subbing:

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

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

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

so 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 $$c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$$

so the even terms are: $$\frac{C_0}{2^n \cdot n!}$$ and the odd terms are: $$\frac{C_1}{1 \cdot 3 \cdot 5 \cdot (2n-1)}$$

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

but I'm stuck here. If $$y(0) = 1$$ ... doesn't the equation become 0? I feel like I've hit an impossible condition so I feel like I've made a mistake somewhere.

• When you have substituted in, you have changed the minimum value of n from 0,1 and 2 respectively to 0 across the board. If you do that then you must account for those finite terms as well when you change to zero – W M Seath May 10 at 15:47
• @WMSeath I don't get it... can you show me what you mean? – Jwan622 May 10 at 15:47
• I'm going from n = 2 to n = 0 so aren't the constants taken into account already? – Jwan622 May 10 at 15:55

You did well. Note that the initial conditions $$y(0)=1$$ and $$y'(0)=0$$ simply means $$C_0=1$$ and $$C_1=0$$.
Also, noting that $$C_1=0$$, your final solution is $$y=\sum_{n=0}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}=c_0+\sum_{n=1}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}$$ which dosn't make the equation $$0$$.
• No, you did every thing right. What you did, is what every body do. The only typo was that when you substituted $x=0$ in the final solution, you missed to note that the series started at $n=0$, which makes $x^{2n}=0$ in the case of $n=0$. – Qurultay May 10 at 16:33
• What I mean: first apply the index $n$, then substitute value of $x$. – Qurultay May 10 at 16:34