Answer :
To avoid a trivial solution of the differential equation y = sin(kt)y'' + 49y = 0, the value of k must be such that sin(kt) is zero only at integral multiples of π. The correct non-trivial solutions for k would be ±7, since these satisfy the differential equation without resulting in a zero function. Therefore option b. k = 7, - 7e
The student is tasked with finding the values of k that ensure the function y does not become trivial (zero everywhere) when it satisfies the differential equation y = sin(kt)y'' + 49y = 0. We seek non-trivial solutions, which generally provide insights into the behavior of a system. Given the sin function is involved in the differential equation, we know from trigonometry that sin(a) = 0 for a being an integral multiple of π. Hence, for a non-trivial solution, the sine function within the differential equation should be zero at only specific points, implying kt must be an integral multiple of π. If kt is considered as an angle in radians, k should be chosen such that this condition is satisfied, and k must not cause the entire function to become zero.
Since the given equation also contains a term 49y, we look for harmony in terms of the trigonometric oscillation of sine with the constant term. Specifically, the equation suggests a relationship to the square of the sine's wave number, k, with the term 49, which is reminiscent of the form seen in sinusoidal solutions to harmonic motion equations. Thus, we are essentially looking for where the oscillatory sin term and the constant term balance out to satisfy the differential equation without yielding a zero function. Considering this interpretation, the wave number k needs to be in harmony with 49, suggesting that k could be ±7, since 7^2 equals 49. This gives us the correct wavelengths to balance the equation without resulting in triviality.
