Differential Equations - Verifying a solution of a given DE

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Homework Help Overview

The discussion revolves around verifying a solution to a differential equation, specifically examining the function \(\ln \frac{2-x}{1-x} = t\) as a potential solution to the equation \(\frac{dX}{dt} = (2-x)(1-x)\). Participants are exploring the relationship between the derivative of the logarithmic function and the given differential equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to differentiate the logarithmic function to verify if it satisfies the differential equation. There are questions about the correct application of differentiation techniques, including the chain rule and properties of logarithms.

Discussion Status

Some participants have suggested taking the derivative of the logarithmic function, while others have expressed uncertainty about the results they obtained. There is an ongoing exploration of the differentiation process and its implications for verifying the solution.

Contextual Notes

One participant notes difficulty in understanding the problem and the relationship between the derivative and the differential equation. There is mention of a solution provided in a textbook, which may not align with the participants' calculations.

Nickg140143
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Homework Statement



Verify that the indicated funciton is a solution of the given Differential Equation. c1 and c2 denote constants where appropriate.

[tex]\frac { dX }{ dt } =(2-x)(1-x);\quad \quad \ln { \frac { 2-x }{ 1-x } } =t[/tex]

The Attempt at a Solution


I'm not quite sure how to really start this problem. If I'm reading the question right, the differential equation is

[tex]\frac { dX }{ dt } =(2-x)(1-x)[/tex]

and that the solution I'm checking is
[tex]\ln { \frac { 2-x }{ 1-x } } =t[/tex]

I was thinking of perhaps integrating
[tex]\frac { dX }{ dt } =(2-x)(1-x)[/tex]
then maybe plug in the the given value of t?

I've read through the section and looked at my notes from class, but I can't seem to fully understand what I should be doing in this problem. The solution in the back of the book says

[tex]\frac { d }{ dt } \ln { \frac { 2-X }{ 1-X } } =1,\quad \left[ \frac { -1 }{ 2-X } +\frac { 1 }{ 1-X } \right] \frac { dX }{ dt } =1[/tex]
Simplifies to
[tex]\frac { dX }{ dt } =(2-X)(1-X)[/tex]

...Not entirely sure what it necessarily means by that though. Any help would be GREATLY appreciated.
 
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I think they just want you to take the derivative of [tex]\ln { \frac { 2-x }{ 1-x } }[/tex] and find that it does indeed equal [tex](2-x)(1-x)[/tex]

EDIT: although when I take the derivative I get the answer upside down. I get [tex]\frac{1}{(1-x)(2-x)}[/tex]
 
Last edited:
You just need to take the derivative of the natural log function. Some things you might want to keep in mind when doing this are that you can split this function into two natural logs, and also the chain rule.
 
Alright, I'll give that a go and try to make sense of the question, thanks for the help you two
 

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