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EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!

In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.

I have noted that $y^n$ refers to the nth derivative.

I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.

That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.

The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.

The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.

Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.

I am not sure if this is relevant in breaking down the proof into cases...

Thank you in advance for any insight.

$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.

We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.

$frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) cdots $ (i)

Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) cdots $ (ii)

Adding equations (i) and (ii) together we indeed get $0$ as the result. Hence, Proved