Q.Form the differential equation representing the family of curves $y=A\cos x+B\sin x$, where $A,B$ are arbitrary constants.

Two arbitrary constants require differentiating twice; the second derivative reproduces the negative of the original function.

$\frac{dy}{dx}=-A\sin x+B\cos x$; $\frac{d^2y}{dx^2}=-A\cos x-B\sin x=-y$.

Differentiating twice (two constants need two differentiations)

Recognising the elimination

The right-hand side of the second derivative is exactly $-(A\cos x+B\sin x)=-y$:

Check (independent recomputation, verifying with the original family directly substituted): for $y=A\cos x+B\sin x$, computing $\frac{d^2y}{dx^2}$ directly gives $-A\cos x-B\sin x$, which is literally $-y$ by the original equation — so $\frac{d^2y}{dx^2}+y=0$ holds for EVERY choice of $A,B$, confirming the family-wide differential equation.

Stopping after only ONE differentiation — with two independent arbitrary constants, a single derivative still contains both $A$ and $B$ and cannot eliminate them; a second differentiation is required.