Chapter 24: Problem 31
The solution of the differential equation, \(\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}\), when \(y(1)=1\), is : (a) \(\log _{\mathrm{e}}\left|\frac{2-x}{2-y}\right|=x-y\) (b) \(-\log _{e}\left|\frac{\mid-x+y}{1+x-y}\right|=2(x-1)\) (c) \(-\log _{e}\left|\frac{1+x-y}{1-x+y}\right|=x+y-2\) (d) \(\log _{e}\left|\frac{2-y}{2-x}\right|=2(y-1)\)
Short Answer
Step by step solution
Separate Variables
Integrate Both Sides
Solve for \( C \) Using Initial Condition
Simplify and Re-evaluate Integration
Substitute and Differentiate the Initial Condition Again
Test Given Options for Validity
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Separation
This means you rearrange the terms such that all \( y \) terms, including \( \mathrm{d} y \), are on one side, and all \( x \) terms, including \( \mathrm{d} x \), are on the other. For this specific equation, you can achieve this by rewriting it as:
- \( \frac{\mathrm{d} y}{(x-y)^2} = \mathrm{d} x \).
Integration
The left side is an integral with respect to \( y \), and the right side is an integral with respect to \( x \).
- The integral of \( \frac{1}{(x-y)^2} \) with respect to \( y \) gives us \( -\frac{1}{x-y} \).
- The integral of \( 1 \) with respect to \( x \) gives us \( x + C \), where \( C \) is the constant of integration.
\[ -\frac{1}{x-y} = x + C \]
Integration is a powerful tool because it transforms a problem of finding a derivative into a problem of computing an anti-derivative, extending the method for solving equations expressed through their rates of change.
Initial Condition
By substituting \( x = 1 \) and \( y = 1 \) into the integrated equation \( -\frac{1}{x-y} = x + C \), we attempt to solve for \( C \):
- \( -\frac{1}{1-1} = 1 + C \)
Initial conditions are essential as they enable us to take a general solution and transform it into a specific solution by solving for the constant.
Constant of Integration
The constant of integration arises because the derivative of a constant is zero in the original function, hence it cannot be determined solely by the integration process and remains undetermined until an initial condition or boundary value is applied.
Re-evaluating based on the given initial condition, you solve \( -\frac{1}{x-y} = x + C \) for \( C \).
- Here, initial attempts to apply \( y(1)=1 \) led to issues primarily due to the setup leading to undefined terms.