Question

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

Short answer

Proved

Step-by-step solution

Step 1. Given

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

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