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Engineering
Mathematics
Variable Separable Differential Equation
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 :

I−k(T−t)22

I−kT22

e – kT

T2−Ik

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Solution

dv(t)dt=k(T−t)

òdv(t) = ò(–kT)dt + òktdt

V(t)=−kTt+kt22+c

at   t = 0   C = I

V(T)=−kTt+kt22+I

Now      at    t = T

V(T)=−kT2+kT22+I

V(T)=I−12kT2

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