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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.

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  • $\begingroup$ 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 $\endgroup$ – W M Seath May 10 at 15:47
  • $\begingroup$ @WMSeath I don't get it... can you show me what you mean? $\endgroup$ – Jwan622 May 10 at 15:47
  • $\begingroup$ I'm going from n = 2 to n = 0 so aren't the constants taken into account already? $\endgroup$ – Jwan622 May 10 at 15:55
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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 $0$.

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  • $\begingroup$ which is fine since y(0) = 1 right? $\endgroup$ – Jwan622 May 10 at 16:28
  • $\begingroup$ the comment from WMseath doesn't apply right? $\endgroup$ – Jwan622 May 10 at 16:29
  • $\begingroup$ Yes, of course. $\endgroup$ – Qurultay May 10 at 16:29
  • $\begingroup$ 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 $n=0$. $\endgroup$ – Qurultay May 10 at 16:33
  • $\begingroup$ What I mean: first apply the index $n$, then substitute value of $x$. $\endgroup$ – Qurultay May 10 at 16:34

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