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Chapter 3: Problem 23

Find \(d y\) $$2 y^{3 / 2}+x y-x=0$$

Short Answer

Expert verified
\(dy = \frac{1 - y}{3y^{1/2} + x} dx.\)

Step by step solution

01

Differentiate the Equation Implicitly

To find the differential \(dy\), we need to differentiate the given equation with respect to \(x\). The original equation is \(2y^{3/2} + xy - x = 0\). Implicit differentiation involves differentiating each term separately with respect to \(x\). For \(2y^{3/2}\), we use the chain rule, resulting in \(3y^{1/2} \cdot dy/dx\). For \(xy\), we use the product rule: \(x\cdot dy/dx + y\). The derivative of \(-x\) is \(-1\). So we get: \[3y^{1/2} \cdot \frac{dy}{dx} + y + x\frac{dy}{dx} - 1 = 0.\]
02

Solve for \(\frac{dy}{dx}\)

Now, we need to collect terms involving \(\frac{dy}{dx}\) to one side of the equation. Rearrange the differentiated equation to get: \[3y^{1/2} \cdot \frac{dy}{dx} + x\frac{dy}{dx} = 1 - y.\] Factor out \(\frac{dy}{dx}\): \[\frac{dy}{dx} (3y^{1/2} + x) = 1 - y.\] Divide both sides by \(3y^{1/2} + x\) to solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{1 - y}{3y^{1/2} + x}.\]
03

Rewrite \(dy\) Using \(\frac{dy}{dx}\)

Having found \(\frac{dy}{dx}\), we can express \(dy\) in terms of \(dx\) as follows: \[dy = \frac{1 - y}{3y^{1/2} + x} dx.\] This expression gives us the change in \(y\) relative to changes in \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. In the context of the given problem, we encounter it when differentiating the term \(2y^{3/2}\). Here, \(y\) is not an explicit function of \(x\), but rather an implicit one, revealing the magic of implicit differentiation.
  • We start with \(2y^{3/2}\), needing the derivative with respect to \(x\).
  • First, differentiate \(y^{3/2}\) in terms of \(y\), yielding \(\frac{3}{2}y^{1/2}\).
  • As \(y\) is a function of \(x\), apply the chain rule: multiply by \(\frac{dy}{dx}\).
By putting this together, the derivative of \(2y^{3/2}\) using the chain rule becomes \(3y^{1/2} \cdot \frac{dy}{dx}\). This aspect of the chain rule helps us manage problems with layers of functions.
Product Rule
The product rule is essential when differentiating product terms, such as \(xy\) in our exercise. This rule dictates how to take the derivative of the product of two separate functions.
For a product \(uv\), the derivative \(\frac{d}{dx}(uv)\) is given by:
  • \(u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\)
In our case:
  • Identify \(u = x\) and \(v = y\)
  • The derivative becomes \(x \cdot \frac{dy}{dx} + y\).
By applying the product rule, we appropriately account for both parts of the product term, setting up for further manipulation and solution.
Differential Equations
Differential equations involve derivatives and describe how a function changes. They are crucial in many fields for modeling systems and understanding change. In this exercise, our goal was to find \(dy\), illustrating a simple form of solving a differential equation.
Here's how we approached it:
  • Started with the implicit equation \(2y^{3/2} + xy - x = 0\).
  • Differentiated using the chain and product rules.
  • Collected terms involving \(\frac{dy}{dx}\) to isolate and solve the differential equation.
The solution \(dy = \frac{1 - y}{3y^{1/2} + x} dx\) describes the relationship between small changes in \(x\) and \(y\). This illustrates how differential equations allow us to track the dynamics of variables in calculus.

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