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I am studying for a discrete mathematics exam and have gotten stuck on this question:

Any function y of a real variable x that solves the diff erential equation:

$$\frac{d^4y}{dx^4} -16y =0$$

may be represented by a power series of the form: $$y = \sum_{n=0}^{+\infty} \frac{y_n}{n!} x^n$$

where the coefficients $y_0, y_1, y_2, y_3,...$ of this power series are real numbers. Find values of these coefficients $y_n$ for $n =0,1,2,3,4,...$ that yields a solution to the above differential equation with $y_0=1,$ $ y_1=0, $ $y_2=-4$ and $y_3=0$.

I have been researching this on the web but have been having trouble with this particular question. Can anyone help? I am still trying to learn this topic so would appreciate any insight into your thought process. Thank you very much. :)

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Why does he ask you for information he already gave? $y_0$, $y_1$, $y_2$, $y_3$ More specifically – Ivan Lerner Aug 23 '13 at 22:15
this is an eigenvalue problem with the general solution $c_1\cosh2x+c_2\sinh2x+c_3\cos2x+c_4\sin2x$ -- it's probably easier to find the series for such an expression rather than solve using series to begin with. @IvanLerner you end up with a fourth-order recurrence and need four initial conditions – oldrinb Aug 23 '13 at 22:17

Hint: $$y^{(4)}(x) = \sum_{n=0}^{+\infty} \frac{y_{n+4}}{n!}x^n.$$

So $y_{n+4} = 16y_n$.

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Hi, sorry for the late reply, I was sleeping. :) I'm not sure as to why the equation didn't change when you differentiated it, would you mind explaining how you got this? Thanks very much for your reply. – KevinH Aug 24 '13 at 7:17
I differentiated the series, not the equation. If $$y(x)=y_0 + y_1 x + \frac{y_2}{2!}x^2 + \frac{y_3}{3!}x^3 + \frac{y_4}{4!}x^4 + \frac{y_5}{5!}x^5 + \ldots$$ then $$y^{(4)}(x)=y_4 + y_5 x + \frac{y_6}{6!}x^2 + \frac{y_7}{7!}x^3 + \ldots.$$ Now all you have to do is to substitute these expressions in the equation. – njguliyev Aug 24 '13 at 8:52

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