Solve Seperable Equation: Step-by-Step Guide

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Homework Help Overview

The discussion revolves around solving a separable differential equation, specifically in the context of integrating functions involving both y and x. Participants are exploring the integration process and the application of techniques such as integration by parts.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration steps involved in solving the equation, questioning the correctness of previous integrations and exploring the application of integration techniques. There is a focus on the integration of terms like y·exp(-y) and the use of integration by parts.

Discussion Status

The conversation is active, with participants providing feedback on integration attempts and suggesting techniques. There is acknowledgment of errors in integration, and some participants express confidence in their understanding of the integration process.

Contextual Notes

Some participants are grappling with the integration of specific terms and the implications of their integration choices. There is an ongoing examination of the assumptions made during the integration process.

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[SOLVED] Seperable Equation

1. Instruction: Solve the equation.

2. Equations:
dy/dx = g(x)p(y)
h(y) = 1/p(y)
h(y)dy = g(x)dx
[tex]\int[/tex]h(y)dy = [tex]\int[/tex]g(x)dx
H(y) = G(x) + C


3. http://img354.imageshack.us/img354/6475/mathdl0.jpg

I tried to do it on the right side, but...I got stuck there. If I add to the right, then I would be left with -C + e[tex]^{-y}[/tex] = ex + -ye[tex]^{-y}[/tex]. Can I say that -C = C? But what about e[tex]^{-y}[/tex] . Did I inegrate it right?

Thank you in advance.
 
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Nope. Didn't integrate it right. Try taking d/dy on the left side. What went wrong?
 
so I should be integrating

[tex]\underline{d}[/tex]= [tex]\underline{d(e^{y})}[/tex]
dx(e[tex]^{x}[/tex]) dy (y-1)
 
You were OK up to the next to the last step. How do you integrate y·exp(-y) ?
 
I multiplied ye[tex]^{-y}[/tex]dy - e[tex]^{-y}[/tex]dy, then integrate
 
sourlemon said:
I multiplied ye[tex]^{-y}[/tex]dy - e[tex]^{-y}[/tex]dy, then integrate

Right, and you integrated the *second* term correctly. What integration technique must you use on the term ye[tex]^{-y}[/tex] ?
 
du dv right?

I think I got it! thank you so much!
 
sourlemon said:
du dv right?

If you mean by that, "integration by parts", we are in agreement. I hope that works out for you...
 

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