Solving the Differential Equation e2xy' + 2e2xy = (6x + 5)e2x

  • Context: Undergrad 
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Discussion Overview

The discussion centers on solving the differential equation \( e^{2x}y' + 2e^{2x}y = (6x + 5)e^{2x} \). Participants explore methods for solving this first-order linear nonhomogeneous equation, including the use of integrating factors and the validity of proposed solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a method involving rearranging the equation and suggests a solution for \( y \).
  • Another participant questions the validity of the proposed method and asks for verification of its correctness.
  • A third participant identifies the equation as a nonhomogeneous first-order linear equation and suggests finding an integrating factor.
  • Subsequent replies confirm that the integrating factor is \( e^{2x} \) and discuss the implications of multiplying the equation by this factor.
  • It is noted that the initial proposed method fails because the integral of \(-2ydx\) does not simplify as suggested, due to \( y \) being a function of \( x \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial proposed solution, with some arguing against its validity and others providing alternative methods. The discussion remains unresolved regarding the effectiveness of the various approaches presented.

Contextual Notes

Limitations include the need for clarity on the assumptions behind the proposed methods and the specific steps involved in integrating the right side of the equation. The discussion does not resolve these mathematical steps.

Icebreaker
[tex]y' + 2y = 6x + 5[/tex]

Can I solve it by...

[tex]\frac{dy}{dx} + 2y = 6x + 5[/tex]

[tex]dy = (6x + 5 - 2y)dx[/tex]

[tex]y = 3x^2 + 5x - 2yx[/tex]

[tex]y + 2yx = 3x^2 + 5x[/tex]

[tex]y = \frac{3x^2 + 5x}{1 + 2x}[/tex]
 
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No you can't.
1) Why do you think this works?
2) Have you checked if it works?
 
Nope.It's a nonhomogenous I-st order linear equation with constant coefficients.Find an integrating factor.

Daniel.
 
Damn...
 
I think it would be [itex]e^{2t} [/tex], right?[/itex]
 
That would be the integrating factor, yes.
 
Well, e2x, actually. [tex]\frac{d(e^{2x}y)}{dx}= 2e^{2x}y+ e^{2x}y'[/tex] so multiplying the entire equation by e2x gives e2xy'+ 2e2xy= (e2xy)'= (6x+5)e2x. Integrating the left side gives e2xy and the right side can be integrated by parts.

By the way, the reason your first method fails is that the integral of -2ydx is not -2xy because y is itself a function of x.
 

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