Chapter 6.3, Problem 31E

To determine

To find: The value of differential equation.

Answer to Problem 31E

The answer is:

  $y=\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\sqrt{{\theta }^2-1}+C$

Explanation of Solution

Given information:

The differential equation is: $\frac{dy}{d\theta }=\theta {\sec }^{-1}\theta ,\theta >1$

Calculation:

The given differential equation is: $\frac{dy}{d\theta }=\theta {\sec }^{-1}\theta ,\theta >1$

Put the variables in a separated row with the $y$ s on the left and the $x$ s on the right. Integrate the left side after writing an integral symbol on both sides:

To use integration by parts $\int udv=uv-\int vdu$

Choose

  $\begin{array}{c}u={\sec }^{-1}\theta \\ dv=\theta d\theta \end{array}$

Then,

  $\begin{array}{c}du=\frac{1}{\left|\theta \right|\sqrt{{\theta }^2-1}}d\theta \\ v=\frac{1}{2}{\theta }^2\end{array}$

Hence,

  $\theta >1,du$ simplifies $du=\frac{1}{\theta \sqrt{{\theta }^2-1}}d\theta$

  $\begin{array}{c}\frac{dy}{d\theta }=\theta {\sec }^{-1}\theta \\ dy=\theta {\sec }^{-1}\theta d\theta \\ \int dy=\int \theta {\sec }^{-1}\theta d\theta \\ y=\int \theta {\sec }^{-1}\theta d\theta \\ ={\sec }^{-1}\theta \left(\frac{1}{2}{\theta }^2\right)-\int \frac{1}{2}{\theta }^2\left(\frac{1}{\theta \sqrt{{\theta }^2-1}}d\theta \right)\\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\int \frac{\theta }{\sqrt{{\theta }^2-1}}d\theta \\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\int \theta {\left({\theta }^2-1\right)}^{-1/2}d\theta \end{array}$

Evaluate the integral to use the substitution

  $\begin{array}{c}u={\theta }^2-1\\ \to du=2\theta d\theta \\ \to \frac{1}{2}du=\theta d\theta \end{array}$

Consider the integral. Considering that it is an indefinite integral, don't forget to write $+C$ :

  $\begin{array}{c}\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\int u^{-1/2}\left(\frac{1}{2}du\right)\\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{4}\int u^{-1/2}du\\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{4}\left(\frac{u^{1/2}}{1/2}\right)+C\\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\sqrt{u}+C\\ =\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\sqrt{{\theta }^2-1}+C\end{array}$

Therefore the required value of the given differential equation $\frac{dy}{d\theta }=\theta {\sec }^{-1}\theta ,\theta >1$ is

  $y=\frac{1}{2}{\theta }^2{\sec }^{-1}\theta -\frac{1}{2}\sqrt{{\theta }^2-1}+C$ .