A.)verify that thr given functions are solutions of the differential equation.

b.) calculate the wronskian. do the two functions form a fundamental set of solutions?

c.) if they form a fundamental set, determine the unique solution of the initial value problem.

1.) y'' - y = 0; y1(t) = -cost, y2(t) = -sint; y(pi/2) = 1, y'(pi/2) = 1

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 )

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²