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

Question

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.

Answer 1

$$y^{''} =\frac {d^2}{dt^2}y= \frac {d}{dt}\left({\frac d{dt}y}\right)= \frac {d}{dt}(-Ak\sin(kt)+Bk\cos(kt))=k\frac {d}{dt}(-A\sin(kt)+B\cos(kt))=k^2(-A\cos(kt)-B\sin(kt))=-k^2y$$ You must have done something erroneous! may be the derivatives of $\sin(x)$ and $\cos(x)$.

Answer 2

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