Chapter 6.4, Problem 11E

To determine

To solve: the given initial value problem by using the separation of variables.

Answer to Problem 11E

  $y=100e^{1.5t}$ .

Explanation of Solution

Given:

  $\frac{dy}{dt}=ky$ , where $k=1.5$ and $y\left(0\right)=100$ .

Concept used:

A differential equation of the form $\frac{dy}{dx}=f\left(y\right)g\left(x\right)$ is called separable.

The variables can be separated by writing it in the form $\frac{1}{f\left(y\right)}dy=g\left(x\right)dx$ .

The solution is found by anti differentiating each side with respect to its thusly isolated variable.

Calculation:

Consider the given differential equation,

  $\frac{dy}{dt}=ky$

Now by separating the variables, it follows that

  $\begin{array}{l}\frac{dy}{y}=kdt\\ \frac{dy}{y}=1.5dt\end{array}$

Now by integrating on both sides, it follows that

  $\begin{array}{l}{\displaystyle \int \frac{dy}{y}}={\displaystyle \int 1.5dt}\\ \ln \left|y\right|=1.5t+C\end{array}$

Now by using the given initial conditions, it follows that

  $\begin{array}{c}\ln 100=1.5\left(0\right)+C\\ C=\ln 100\end{array}$

Now put $C=\ln 100$ in $\ln \left|y\right|=1.5t+C$ , it follows that

  $\begin{array}{c}\ln \left|y\right|=1.5t+\ln 100\\ \ln \left|y\right|-\ln 100=1.5t\\ \ln \left|\frac{y}{100}\right|=1.5t\\ \frac{y}{100}=e^{1.5t}\\ y=100e^{1.5t}\end{array}$