Chapter 4.2, Problem 7QR

To determine

To solve: The derivative of y with respect to x in terms of y and x.

Answer to Problem 7QR

  $y\text{'}=\frac{y+x\times \cos x}{\sin x-x}$

Explanation of Solution

Given information: The equation is given as,

  $y\text{'}\sin x-x\cos x=xy\text{'}+y$

Calculation: Rearranging the equation in terms of y’ ,

  $\begin{array}{l}y\text{'}\sin x-x\cos x=xy\text{'}+y\\ y\text{'}\times \sin x-xy\text{'}=y+x\times \cos x\\ y\text{'}\times \left(\sin x-x\right)=y+x\times \cos x\\ y\text{'}=\frac{y+x\times \cos x}{\sin x-x}\end{array}$

So the derivative of y with respect to x is $y\text{'}=\frac{y+x\times \cos x}{\sin x-x}$