Undergrad Y'' + y = 0 solution and recursion relation

[Sam D]

I've found the general solution to be y(x) = C1cos(x) + C2sin(x).

I've also found a recursion relation for the equation to be:

An+2 = -An / (n+2)(n+1)

I now need to show that this recursion relation is equivalent to the general solution. How do I go about doing this?

Any help would be greatly appreciated!

 

[hilbert2]

So is this the recurrence relation for the coefficients $c_i$ in the power series representing the solution: $y(x)=c_0 + c_1 x + c_2 x^2 \dots$ ? It shouldn't be difficult to take the sum of the power series of sine and cosine multiplied by a constant, and then show that the result satisfies both the DE and that recurrence relation.

 

[vela]

Try calculating the first few terms of the series in terms of $a_0$ and $a_1$. Hopefully, you'll recognize the resulting two series.

 

[SACHIN KUMAR]

<Moderator's note: Approved as it is more than two weeks since the OP has been seen. Member has been warned not to post full solutions. This is an exception as it closes the thread.>

Let's start from your reccurence relation:

$A_{n+2}=\frac{-A_N}{(n+2)(n+1)}$

First collect even $n$ values:

$A_2= \frac{-A_0}{2.1}$

$A_4= \frac{-A_2}{4.3}=\frac{A_0}{4.3.2.1}=\frac{A_0}{4!}$

$A_6= \frac{-A_4}{6.5}=\frac{-A_0}{6.5.4.3.2.1}=\frac{-A_0}{6!}$

...

Now take odd n values:

$A_3= \frac{-A_1}{3.2}$;

$A_5= \frac{-A_3}{5.4}=\frac{A_1}{5.4.3.2.1}=\frac{A_0}{5!}$

$A_7= \frac{-A_5}{7.6}=\frac{-A_0}{7.6.5.4.3.2.1}=\frac{-A_0}{7!}$

...

So final solution isOn putting the values of $A_n$ in Maclaurin series solution ($y(x)=\sum_{n=0}{ A_n x^n}$),$y(x)=\sum_{n=0}{ \frac{(-1)^n x^{2n}}{ 2n!} + \frac{(-1)^n x^{2n+1}}{ 2n+1!}}$

$y(x)= A_0 \cos (x) + A_1 \sin (x)$