Chapter 7.3, Problem 31E

To determine

To solve the differential equation: $\frac{dy}{d\theta }=\theta {\sec }^{-1}(\theta )$

Answer to Problem 31E

The solution of the differential equation $\frac{dy}{d\theta }=\theta {\sec }^{-1}(\theta )$ is $\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{2}\sqrt{{\theta }^2-1}\right]+C$

Explanation of Solution

Given information: $\frac{dy}{d\theta }=\theta {\sec }^{-1}(\theta )$

Formula used: ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left(u^{\prime }{\displaystyle \int vdx}\right)}}}dx$

                           $\frac{d}{d\theta }\left({\sec }^{-1}x\right)=\frac{1}{x\sqrt{x^2-1}}$

Calculation :

The given equation becomes as:

  $dy=\theta {\sec }^{-1}(\theta )d\theta$

Integrating above equation with respect to $x$ , the expression becomes as:

  $y={\displaystyle \int \theta {\sec }^{-1}(\theta )}d\theta +C$

  $=\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-{\displaystyle \int \left(\frac{1}{\theta \sqrt{{\theta }^2-1}}\right)}\left(\frac{{\theta }^2}{2}\right)d\theta \right]+C$

  $=\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{2}{\displaystyle \int \left(\frac{\theta }{\sqrt{{\theta }^2-1}}\right)}d\theta \right]+C$

Put ${\theta }^2-1=t$ , so that $2\theta d\theta =dt$ , the above expression becomes as:

  $=\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{4}{\displaystyle \int \frac{1}{\sqrt{t}}}dt\right]+C$

  $=\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{2}\sqrt{t}\right]+C$

  $=\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{2}\sqrt{{\theta }^2-1}\right]+C$

Hence, the solution of the differential equation is $\left[{\sec }^{-1}(\theta )\left(\frac{{\theta }^2}{2}\right)-\frac{1}{2}\sqrt{{\theta }^2-1}\right]+C$