Let y₁ be a solution to
y'' + P(x)y' + Q(x)y = 0, x ∈ I
such that y₁(x₀) = y₁'(x₀) = 0, for some x₀ ∈ I, where P(x) and Q(x) are continuous on I. Then which of the following is/are TRUE?
(a) y₁ is linearly dependent to any other solution y₂ of the differential equation.
(b) y₁ = 0.
(c) W(x) = 0, ∀ x ∈ I, where W(x) is the Wronskian of y₁ and any other solution y₂.
(d) All of the above.
Answer (1)
Let's analyze the given differential equation and the specific scenario described:
y ′′ + P ( x ) y ′ + Q ( x ) y = 0 ,
where y 1 is a solution such that y 1 ( x 0 ) = y 1 ′ ( x 0 ) = 0 for some x 0 . We need to determine which of the options are true based on this information.
Step-by-Step Analysis:
Option (a): y₁ is linearly dependent to any other solution y₂ of the differential equation.
According to the uniqueness theorem for linear differential equations, if a non-trivial solution exists which satisfies these conditions, then this solution must be identically zero for all x in I . Therefore, any non-zero solution of the differential equation must be linearly dependent on a non-trivial y 1 because y 1 cannot be non-trivial.
Option (b): y₁ = 0.
Since the only solution that can satisfy the conditions y 1 ( x 0 ) = y 1 ′ ( x 0 ) = 0 throughout an interval is the trivial solution, y 1 ( x ) = 0 for all x ∈ I . Thus, this option is true.
Option (c): W(x) = 0, ∀ x ∈ I, where W(x) is the Wronskian of y₁ and any other solution y₂.
The Wronskian W ( y 1 , y 2 ) = y 1 y 2 ′ − y 2 y 1 ′ . Since y 1 = 0 and y 1 ′ = 0 for all x , it follows that W ( x ) = 0 for all x ∈ I regardless of y 2 . Therefore, this option is true.
Option (d): All of the above.
Given that each of the previous options (a), (b), and (c) are true based on the conditions described for y 1 , this option is also true.
Conclusion:
The correct answer is (d) All of the above. Each statement follows logically from the properties and theorems applicable to solutions of linear homogeneous differential equations with continuous coefficients.
