Solving D.eq. with Homogeneous Method: Tips and Tricks for Beginners

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around solving a differential equation using the homogeneous method. Participants explore various strategies and techniques for beginners, focusing on the transformation of variables and the handling of constants in the equation.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant expresses difficulty in solving the differential equation using the homogeneous method and seeks alternative methods.
  • Another participant suggests solving the simultaneous equations derived from the differential equation and proposes a change of variables to simplify the problem.
  • A later reply notes that after applying the change of variables, the constants disappear, prompting a question about how to address the remaining side of the differential equation.
  • Another participant advises differentiating both sides after the variable transformation to relate the new dependent variable to the original one.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus, as participants present different approaches and raise questions about the process without resolving the issues raised.

Contextual Notes

Participants do not clarify certain assumptions regarding the simultaneous equations or the specific nature of the differential equation, leaving some steps and dependencies unresolved.

Who May Find This Useful

Beginners in differential equations, particularly those interested in the homogeneous method and variable transformations.

em919
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I need some help to solve this D.eq. I've tried the homogeneous method but I didn't work!

Which method can I use?

y'=(-3x+4y-18)/(2x-y+7)
 
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Hi, em919! ;)

First of all, you need to solve the simultaneous equations

[tex]-3x+4y-18 = 0,[/tex]

[tex]2x-y+7 = 0.[/tex]​

If the solution to this set of simultaneous equations is [tex](x_1, y_1)[/tex], you need to make a couple of changes of variables: [tex]t = x + x_1[/tex] and [tex]z = y + y_1[/tex]. Your new independent variable will be [tex]t[/tex] and your new dependent variable will be [tex]z[/tex]; you'll see that this new differential equation in [tex]z[/tex] is homogeneous. Once you solve this equation, you need to plug back into get the solution in terms of [tex]y[/tex] and [tex]x[/tex].

Hope this helps. :)
 
Nice!
 
When I do this the constants go away but what should I do with the other side of the d.eq?
 
Differentiate both sides in [tex]z = y + y_1[/tex]; you'll see that [tex]z' = y'[/tex].
 

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