# Given $y = A\cos(kt) + B\sin(kt)$, where $A, B, k$ are constants, prove that $y^{(2)} + (k^2)y = 0$

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, $$y^n = -Ak^n\sin kt + Bk^n\cos kt$$.

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.

• I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative) – WW1 Dec 3 '18 at 3:01
• Ah! In that case, I just find the second derivative and sub in! – user424712 Dec 3 '18 at 3:02

$$y^{''} =\frac {d^2}{dt^2}y= \frac {d}{dt}\left({\frac d{dt}y}\right)= \frac {d}{dt}(-Ak\sin(kt)+Bk\cos(kt))=k\frac {d}{dt}(-A\sin(kt)+B\cos(kt))=k^2(-A\cos(kt)-B\sin(kt))=-k^2y$$ 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 $$t$$.
$$\frac{dy}{dt}=-Ak \sin(kt)+Bk \cos(kt) \implies \frac{d^2y}{dt^2}=-Ak^2 \cos(kt)+Bk^2 \sin(kt) \tag1$$
Also, $$k^2y=Ak^2 \cos(kt)+Bk^2 \sin(kt) \tag2$$
Adding equations $$(1)$$ and $$(2)$$ together we indeed get $$0$$ as the result. Hence, Proved