Chapter 6.3, Problem 13E

To determine

To find: The initial equation.

Answer to Problem 13E

The initial equation: $u=x\tan x+\ln \left|\cos x\right|+1$

Explanation of Solution

Given information:

The differential equation is: $\frac{du}{dx}=x{\sec }^{2}x$

And $u=1$ when

  $x=0$

Calculation:

The given differential equation is: $\frac{du}{dx}=x{\sec }^{2}x$

Integrate each side after separating the variables. Use part-by-part integration

  $\begin{array}{c}\frac{du}{dx}=x{\sec }^{2}x\\ du=x{\sec }^{2}xdx\\ \int du=\int x{\sec }^{2}xdx\\ u=x\tan x-\int \tan xdx\\ =x\tan x-(-\ln |\cos x|)+C\\ =x\tan x+\ln \left|\cos x\right|+C\end{array}$

To find $C$ , use the initial conditions.

And $u=1$ when

  $x=0$

  $\begin{array}{c}u=x\tan x+\ln \left|\cos x\right|+C\\ 1=0\tan 0+\ln \left|\cos 0\right|+C\\ 1=0+\ln 1+C\\ 1=0+C\\ 1=C\end{array}$

The equation is then $u=x\tan x+\ln \left|\cos x\right|+1$

The slope field and the function's graph do match.

  

Therefore the required initial equation of given differential equation $\frac{du}{dx}=x{\sec }^{2}x$ is

  $u=x\tan x+\ln \left|\cos x\right|+1$ .