The characteristic equation for the differential equation y′′ − y = 0 is r^2 - 1 = 0. This has roots r = ±1. Therefore, the general solution to the differential equation is y(t) = c1 e^t + c2 e^(-t).
Using the initial conditions, we can solve for the values of c1 and c2:
y(0) = 5/4 = c1 + c2
y'(0) = -3/4 = c1 - c2
Solving these equations simultaneously, we get c1 = 1/2 and c2 = 3/4. Therefore, the solution to the initial value problem is:
y(t) = 1/2 e^t + 3/4 e^(-t)
To find the minimum value of the solution, we can take the derivative of y(t) and set it equal to 0:
y'(t) = 1/2 e^t - 3/4 e^(-t)
y'(t) = 0 when e^t = (3/2) e^(-t)
e^(2t) = 3/2
t = ln(sqrt(3/2))
Substituting this value of t back into the original equation, we get the minimum value of the solution:
y(ln(sqrt(3/2))) = 1/2 e^(ln(sqrt(3/2))) + 3/4 e^(-ln(sqrt(3/2)))
y(ln(sqrt(3/2))) = sqrt(3)/4 + 3/(4sqrt(3))
y(ln(sqrt(3/2))) = (4sqrt(3) + 3)/(4sqrt(3))
Therefore, the minimum value of the solution is (4sqrt(3) + 3)/(4sqrt(3)) which is approximately 1.068.
To plot the solution for 0 ≤ t ≤ 2, we can use a graphing calculator or software to graph y(t) = 1/2 e^t + 3/4 e^(-t) and then set the viewing window to show the interval [0, 2].