Solveeit Logo

Question

Mathematics Question on Differential equations

Let II be the purchase value of an equipment and V(t)V(t) be the value after it has been used for tt years. The value V(t)V(t) depreciates at a rate given by differential equation dV(t)dt=k(Tt),\frac{d V(t)}{d t}=-k(T-t), where k>0k>0 is a constant and TT is the total life in years of the equipment. Then the scrap value V(T)V(T) of the equipment is

A

ekTe^{-k T}

B

T21kT^{2}-\frac{1}{k}

C

IkT22I-\frac{k T^{2}}{2}

D

Ik(Tt)22I-\frac{k(T-t)^{2}}{2}

Answer

IkT22I-\frac{k T^{2}}{2}

Explanation

Solution

lV(T)dV(t)=t=0Tk(Tt)dt\int\limits_{l}^{V(T)} d V(t)=\int_{t=0}^{T}-k(T-t) d t V(T)I=k[(Tt)22]0T\Rightarrow \quad V(T)-I=k\left[\frac{(T-t)^{2}}{2}\right]_{0}^{T} V(T)I=k[T22]\Rightarrow \quad V(T)-I=-k\left[\frac{T^{2}}{2}\right] V(T)=IkT22\Rightarrow \quad V(T)=I-\frac{k T^{2}}{2}