Let I be the purchase value of an equipment a | vedclass

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Question

(A) Given the differential equation: $\frac{dV}{dt} = -k(T - t)$.Integrating both sides with respect to $t$:$V(t) = \int -k(T - t) dt = k \int (T - t)

Vedclass · Accepted Answer

$I - \frac{kT^2}{2}$

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(A) Given the differential equation: $\frac{dV}{dt} = -k(T - t)$.Integrating both sides with respect to $t$:$V(t) = \int -k(T - t) dt = k \int (T - t) d(T - t) = \frac{k(T - t)^2}{2} + C$.At $t = 0$,the value of the equipment is the purchase value $I$,so $V(0) = I$:$I = \frac{k(T - 0)^2}{2} + C \implies I = \frac{kT^2}{2} + C \implies C = I - \frac{kT^2}{2}$.Substituting $C$ back into the equation for $V(t)$:$V(t) = \frac{k(T - t)^2}{2} + I - \frac{kT^2}{2}$.The scrap value $V(T)$ is the value at $t = T$:$V(T) = \frac{k(T - T)^2}{2} + I - \frac{kT^2}{2} = 0 + I - \frac{kT^2}{2} = I - \frac{kT^2}{2}$.

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