Q.(a) If $y = a\,\cos mx + b\,\sin mx$, then show that $y_2 + m^{2}y = 0$.

(a) $y_2=-m^{2}y\Rightarrow y_2+m^{2}y=0$. (b) Base $n=1$ holds; assuming $k^{2}+k$ even gives $(k+1)^{2}+(k+1)$ even.

Part (a) — Second-order derivative.

Given $y=a\cos mx+b\sin mx$.

Step 1 — First derivative (chain rule).

$y_1=\dfrac{dy}{dx}=-am\sin mx+bm\cos mx.$

Step 2 — Second derivative.

$y_2=\dfrac{d^{2}y}{dx^{2}}=-am^{2}\cos mx-bm^{2}\sin mx=-m^{2}\left(a\cos mx+b\sin mx\right).$

Step 3 — Substitute $y$.

$y_2=-m^{2}y\;\Rightarrow\;y_2+m^{2}y=0.$

Part (b) — Mathematical induction: $n^{2}+n$ is even for all $n\in N$.

Let $P(n):\ n^{2}+n\ \text{is even}$.

Step 1 — Base case ($n=1$).

$1^{2}+1=2$, which is even. So $P(1)$ is true.

Step 2 — Inductive hypothesis.

Assume $P(k)$ is true, i.e. $k^{2}+k=2m$ for some integer $m$.

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