Answer:
The given functions are solutions of the differential equation.
W = 1
y(t) = - cos(t) + sin(t
)
Step-by-step explanation:
Given
y'' - y = 0
y₁(t) = - Cos(t)
y₁’(t) = sin(t)
y(π/2) = 1
y'(π/2) = 1
y₂(t) = - sin(t)
y₂'(t) = - cos(t)
a)
y(t) = C₁* y₁(t) + C₂* y₂(t) = - C₁*cos(t) - C₂*sin(t
)
y(π/2) = - C₁*cos(π/2) - C₂*sin(π/2) = - C₂ = 1
⇒ C₂ = - 1
y’(t) = C₁*Sin(t) - C₂*Cos(t)
y’(π/2) = C₁*Sin (π/2) - (-1)*Cos (π/2) = C₁ = 1
⇒ C₁ = 1
y’’(t) = Cos(t) - Sin(t
)
We can verify that y₁(t) and y₂(t) are indeed solutions to Equation by substitution
y'' - y = (Cos(t) - Sin(t)) - (- cos(t) + sin(t)) = 2*cos(t) - 2*sin(t) = cos(t) - sin(t) = 0
They are solutions of the differential equation only if
cos(t) = sin(t) ⇔ t = n*π/4 / n ∈ Z).
b) We compute the Wronskian as follows
W = y₁(t)*y₂'(t) - y₂(t)*y₁’(t)
W = (-cos(t))*( -cos(t)) - (-sin(t))*(sin(t)) = Cos²(t) + sin²(t) = 1 ≠ 0
Then, the two functions form a fundamental set of solutions.
c) y(t) = - cos(t) + sin(t
)
