/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 A singular solution of a differe... [FREE SOLUTION] | 91Ó°ÊÓ

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A singular solution of a differential equation is a solution that cannot be derived from the general solution of the differential equation. Use implicit differentiation to show that \(-1 / x+2 / x^{2}+1 / y-1 / y^{2}=C\) is a general (implicit) solution of the differential equation \(d y / d x=(x-4) y^{3} /\left[x^{3}(y-2)\right]\). Is \(y=0\) a solution of this differential equation? Is \(y=0\) a singular solution?

Short Answer

Expert verified
The implicit solution is verified. Yes, \(y=0\) is a solution but a singular solution.

Step by step solution

01

- Differentiate Implicitly

Differentiate \(-\frac{1}{x}+\frac{2}{x^{2}}+\frac{1}{y}-\frac{1}{y^{2}}=C\) implicitly with respect to \(x\). Remember to apply the chain rule accordingly.
02

- Differentiate Each Term

Calculate the derivatives of each individual term: \(-\frac{d}{dx} \left( \frac{1}{x} \right) + \frac{d}{dx} \left( \frac{2}{x^2} \right) + \frac{d}{dx} \left( \frac{1}{y} \right) - \frac{d}{dx} \left( \frac{1}{y^2} \right) = 0\). For the terms with \(y\), use the chain rule considering \( \frac{dy}{dx} \).
03

- Apply Derivatives and Chain Rule

Apply the derivatives and simplify: \(\frac{1}{x^2} - \frac{4}{x^3} - \frac{1}{y^2} \frac{dy}{dx} + \frac{2}{y^3} \frac{dy}{dx} = 0\)
04

- Solve for dy/dx

Isolate \( \frac{dy}{dx} \) to find: \(\frac{dy}{dx} = \frac{\frac{1}{x^2} - \frac{4}{x^3}}{ \frac{1}{y^2} - \frac{2}{y^3}} = \frac{x^2-4x}{x^3(y-2)} = \frac{(x-4)y^3}{x^3(y-2)} \).
05

- Verifying y=0

Check whether \(y = 0\) is a solution by substituting \(y = 0\) into the differential equation \( \frac{dy}{dx} = \frac{(x-4)y^3}{x^3(y-2)} \). Notice the right-hand side is 0.
06

- Singular Solution Verification

Since \(y = 0\) gives \(\frac{dy}{dx} = 0\), it satisfies the differential equation but does not follow from the general solution where \( \frac{1}{y} - \frac{1}{y^2} \) terms are undefined.

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique used when dealing with equations where the dependent variable \(y\) is not isolated on one side of the equation. For example, consider the equation:
\( -\frac{1}{x} + \frac{2}{x^2} + \frac{1}{y} - \frac{1}{y^2} = C \).
To find the derivative \( \frac{dy}{dx} \), we differentiate each term with respect to \(x\).

Why Use Implicit Differentiation?

  • It helps when \(y\) can't be easily isolated.
  • Allows for differentiation of complex relationships between variables.

When differentiating terms involving \(y\), apply the chain rule. Let's look at two terms separately:
  • \( \frac{d}{dx} \big(\frac{1}{y} \big) = -\frac{1}{y^2} \frac{dy}{dx} \)
  • \( \frac{d}{dx} \big( -\frac{1}{y^2} \big) = 2 \frac{1}{y^3} \frac{dy}{dx} \)
Notice that both derivatives involve \( \frac{dy}{dx} \). This is a key feature of implicit differentiation.
Chain Rule
The chain rule is essential when differentiating composite functions. It states:

  • If \(u = g(x)\) and \(y = f(u)\), then \( \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} \).

Using the Chain Rule in Implicit Differentiation

The chain rule is applied whenever we differentiate an expression that is a function of another function. In the problem, terms involving \(y\) require the chain rule.
For example, in the term \( \frac{1}{y} \), \(y\) itself is a function of \(x\):
  • Find the derivative of \( \frac{1}{y} \) with respect to \(y\).
  • Multiply by \( \frac{dy}{dx} \) to account for the fact \(y\) is a function of \(x\).

Putting it all together:
\( \frac{d}{dx} \big( \frac{1}{y} \big) = -\frac{1}{y^2} \frac{dy}{dx} \)
Differential Equations
A differential equation is an equation that involves derivatives of a function. It describes how a function changes and can often model real-world systems.
Given the equation:
\( \frac{dy}{dx} = \frac{(x-4)y^3}{x^3(y-2)} \)
This is a first-order differential equation because it involves the first derivative \( \frac{dy}{dx} \). Our task is to solve this equation.

General Solutions vs. Singular Solutions

  • A general solution involves arbitrary constants and represents a family of curves.
  • A singular solution is a specific solution that is not obtainable from the general solution and often serves as a boundary. In this case, \(y=0\) is being evaluated to see if it meets these criteria.

To determine if \(y=0\) is a solution, we substitute it into the differential equation and check:
  • If the equation holds, \(y=0\) is a solution.
  • If it satisfies the differential equation but not the general solution's form, it may be a singular solution.

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