Question:

Let I be the purchase value of an equipment and v(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation dV(t)/dt = −k(T−t), where k>0 is a constant and T is the total life in years of the equipment. Then the scrap value v(T) of the equipment is

T²k

I−kT²/2

I−k(T−t)²/2

e⁻ᵏᵀ

dV(t)/dt = -k(T - t)

Integrating both sides with respect to t:
∫dV(t) = ∫-k(T - t)dt
V(t) = -k(Tt - t²/2) + C

At t = 0, V(0) = I (initial value)
I = -k(T(0) - 0²/2) + C
C = I

Therefore, V(t) = -k(Tt - t²/2) + I

At t = T (scrap value):
v(T) = V(T) = -k(T² - T²/2) + I
v(T) = -kT²/2 + I
v(T) = I - kT²/2