Series solution to $y''+(\cos t) y=0$

Question

Series solution to $\mathbf{y''+(cost)y=0}\;$ is$$\sum_{n=0}^{\infty}a_nt^n$$Find $a_2/a_4$

choices are -6, -4, 4, 6

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I found $a_{k+2}=-\frac{\cos t}{(k+1)(k+2)}a_k$
So, $a_4=-\frac{\cos t}{3\cdot4}a_2 \;\to\;a2/a4=\frac{-12}{\cos t}$

I'm not sure how to pick answer from this.
Is information given in this problem enough to solve it?

Answer 1

Hint

We have that

$$y(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4+a_5t^5+a_6t^6+\dots$$ from where

$$y''(t)=2a_2+6a_3t+12a_4t^2+20a_5t^3+30a_6t^4+\dots$$

Since

$$\cos t=1-\dfrac{t^2}{2}+\dfrac{t^4}{24}+\dots $$ it is

$$2a_2+6a_3t+12a_4t^2+20a_5t^3+30a_6t^4+\dots=-\left(1-\dfrac{t^2}{2}+\dfrac{t^4}{24} +\dots\right)\left(a_0+a_1t+a_2t^2+a_3t^3+a_4t^4+\dots\right)$$

In particular, we get

$$2a_2=-a_0, 6a_3=-a_1, 12a_4=-a_2+\dfrac{a_0}{2}.$$