Q. 75 - Calculus

Question

 Prove Theorem 6.20 by solving the initial-value problem $\frac{dQ}{dt}=kQ with Q\left(0\right)=Q_0$, where k is a constant

Step-by-Step Solution

Answer

Proved

Step 1. Given

The given problem is $\frac{dQ}{dt}=kQ with Q\left(0\right)=Q_0$.

Step 2. Proof

$$\begin{gathered} \int \frac{dQ}{Q}=\int kdt \\ \ln \left|Q\right|=kt+C \\ Q=e^{kt+C} \\ =A{e}^{kt} \\ \mathrm{Now} \mathrm{use} \mathrm{the} \mathrm{initial} \mathrm{condition} \mathrm{Q}={\mathrm{Q}}_0\mathrm{for} \mathrm{t}=0 \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{expression} \mathrm{to} \mathrm{obtain} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{initial}-\mathrm{value} \mathrm{problem} \mathrm{as} \\ {Q}_0=A \\ \mathrm{Substitute} \mathrm{this} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{constant} \mathrm{A} \mathrm{in} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{to} \mathrm{obtain} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{initial} - \mathrm{value} \mathrm{problem} \mathrm{as} \\ Q\left(t\right)=Q_0{e}^{kt} \end{gathered}$$Observe that the differential equation does not contain the independent variable at all. So, solve the differential equation by using antidifferentiation method