- #1

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[tex]y'' - y = 0, x_0 = 0[/tex]

Substituting the series for y and y-double prime:

[tex]\sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} - \sum_{n = 0}^{\infty} a_n x^n = 0[/tex]

Now, substituting n + 2 for n in first term:

[tex]\sum_{n = 0}^{\infty} (n + 2)(n + 1) a_{n + 2} x^n - \sum_{n = 0}^{\infty} a_n x^n = 0[/tex]

Combining the series and factoring out x^n:

[tex]\sum_{n = 0}^{\infty} \left[(n + 2)(n + 1) a_{n + 2} - a_n\right] x^n = 0[/tex]

Arriving at the recursion formula:

[tex]a_{n + 2} = \frac{a_n}{(n + 2)(n + 1)}[/tex]

__This is where I am completely stuck.__

*What is it I am supposed to assume for ao and a1? I know they are arbitrary, but what does that mean and how does it help me to solve the differential equation?*

The answer given is:

[tex]y_1(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n}}{(2n)!} = cosh x[/tex]

[tex]y_2(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ... = \sum_{n = 0}^{\infty} \frac{x^{2n + 1}}{(2n + 1)!} = sinh x[/tex]