Chapter 4.2, Problem 6QR

To determine

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

Answer to Problem 6QR

  $y\text{'}=\frac{4}{x}-\frac{y}{x^2}+\frac{2y}{x}$

Explanation of Solution

Given information: The equation is given as,

  $x^{2}y\text{'}-2xy=4x-y$

Calculation: Rearranging the equation in terms of y’ ,

  $\begin{array}{l}{x}^{2}y\text{'}-2xy=4x-y\\ x^{2}y\text{'}=4x-y+2xy\\ y\text{'}=\frac{4x-y+2xy}{x^2}\\ y\text{'}=\frac{4}{x}-\frac{y}{x^2}+\frac{2y}{x}\end{array}$

So the derivative of y with respect to x is $y\text{'}=\frac{4}{x}-\frac{y}{x^2}+\frac{2y}{x}$