Chapter 6, Problem 27RE

To determine

To solve: the given initial value problem.

Answer to Problem 27RE

  $y=\ln \left|t+4\right|+2$ .

Explanation of Solution

Given:

  $\frac{dy}{dt}=\frac{1}{t+4},y\left(-3\right)=2$ .

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,

  $\frac{dy}{dt}=\frac{1}{t+4}$

By separating the variables, it follows that

  $dy=\frac{1}{t+4}dt$

Now by integrating on both sides, it follows that

  $\begin{array}{c}{\displaystyle \int dy}={\displaystyle \int \frac{1}{t+4}dt}\\ y=\ln \left|t+4\right|+C\end{array}$

Now by using the given initial condition, it follows that

  $\begin{array}{c}2=\ln \left|-3+4\right|+C\\ 2=\ln \left(1\right)+C\\ 2=0+C\\ C=2\end{array}$

Now put $C=2$ in $y=\ln \left|t+4\right|+C$ , it follows that

  $y=\ln \left|t+4\right|+2$