Chapter 7.2, Problem 70E

To determine

To show that $y=\ln \left|\frac{\sin x}{\sin 2}\right|+6$ is the solution of the initial value problem $\frac{dy}{dx}=\cot (x),y(2)=6$

Explanation of Solution

Given information: $\frac{dy}{dx}=\cot (x),y(2)=6$

Formula used: ${\displaystyle \int \cot xdx=\ln (\sin x)+C}$

Proof :

  $dy=\cot (x)dx$

Integrating with respect to x , the expression becomes as:

  $y=\ln (\sin (x))+C$

Using, $y(2)=6$

  $6=\ln (\sin (2))+C$

  $\Rightarrow C=6-\ln (\sin (2))$

The expression is $y=\ln (\sin (x))+6-\ln (\sin (2))$

  $y=\ln \left|\frac{\sin x}{\sin 2}\right|+6$

Hence, $y=\ln \left|\frac{\sin x}{\sin 2}\right|+6$ is the solution of $\frac{dy}{dx}=\cot (x)$