We have the ODE:

$$y' = -3y$$

Assume that the solution of the ODE can be written as a Taylor series in the form:

$$y(x) = a_0 + a_1x + a_2x^2 + \ldots + a_nx^n + \ldots$$

Using this infinite series, find an equation that relates $a_{n+1}$ to $a_n$

From that result, deduce the coefficients $a_1, a_2, a_3, \ldots, a_n$ in terms of $a_0$.

I have tried substituting $y(x)$ as well as the derivative of $y(x)$ into the ODE but am unsure what to do next to relate $a_{n+1}$ to $a_n$.

Any help would be greatly appreciated.


I think the answer to the first part is (correct me if I'm wrong) :

$$a_{n+1} = -\frac{3a_n}{n+1}$$

I then used this to find the first few terms ($a_1,a_2,a_3$ etc.) but can't seem to figure out the second part of the problem (deduce the coefficients $a_n$ in terms of $a_0$)

  • $\begingroup$ Welcome to Math.SE. Can you include your substitution of the series into the ODE? $\endgroup$ – Hrodelbert Oct 13 '15 at 8:26
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    $\begingroup$ This is a functional equation: it must hold for any $x$. Can you see a way to use this? $\endgroup$ – Hrodelbert Oct 13 '15 at 9:24
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    $\begingroup$ Okay, have had a think and came up with this: a(n+1) = -3*a(n) / (n+1) $\endgroup$ – Pablo.m Oct 13 '15 at 10:39
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    $\begingroup$ Write your relation to derive expressions for the first few $n$ and see whether you see a pattern emerge that you can use. $\endgroup$ – Hrodelbert Oct 13 '15 at 12:06
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    $\begingroup$ Compute the first few terms and notice the pattern: $a_2 = \frac{(-3)^2}{2!}a_0~~$, $a_3 = \frac{(-3)^3}{3!}a_0~~$, $a_4 = \frac{(-3)^4}{4!}a_0$, $\ldots$. $\endgroup$ – Winther Oct 13 '15 at 23:14

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