Correct Option: A

(i)	dt(t)	+ ty(t) = x(t)
	dt

F[x1(t) + x2(t)] =	dy1(t)	+	dy2(t)	+ t[y1(t) + y2(t)]
	dt		dt

F[x1(t)] + F[x2(t)] =	dy1(t)	+	dy2(t)	+ t[y1(t) + y2(t)]
	dt		dt

Here, F[x₁(t) + x₂(t)] = F[x₁(t)] + F[x₂(t)]
Hence, the system is linear.

(ii)	d2y(t)	+ y(t)	dy(t)	+ y(t) = x(t)
	dt2		dt

F[x1(t)] =	d2	y1(t) + y1(t)	dy1(t)	+ y1(t)
	dt2		dt

F[x2(t)] =	d2	y2(t) + y2(t)	dy2(t)	+ y2(t)
	dt2		dt

Therefore,

aF[x1(t)] + bF[x1(t)] =	d2y1(t)	+ b	d2y2(t)
	dt2		dt2

+ ay1(t)	dy1(t)	+ by2(t)	dy2(t)	+ ay1(t) + by2(t)
	dt		dt

aF[ax1(t) + bx2(t) =	d2	[ay1(t) + by2(t)] + [ay1(t)
	dt2

+ by2(t)]			d	y1(t) +	d	y2(t)		+ by2 [ay1(t) + by2(t) + 1]
			dt		dt

Since here,
aF[x₁(t)] + bF[x₂(t)] ≠ F[ax₁(t) + bx₂(t)]
Hence, the system is non-linear.