Differentials with linear equations

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Homework Help Overview

The discussion revolves around solving a differential equation of the form y' = y(1-x) with an initial condition y(1) = e. Participants are exploring different methods to integrate and solve for y at x = 2.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various integration techniques, including separating variables and integrating both sides. There are attempts to derive expressions for y, but some participants express confusion over the integration process and the implications of the initial condition.

Discussion Status

The conversation is ongoing, with participants sharing different approaches and questioning the correctness of their integration steps. Some guidance has been offered regarding integration methods, but no consensus has been reached on a viable solution.

Contextual Notes

Participants note issues with the initial condition leading to undefined expressions and question the validity of their integration techniques. There is a focus on ensuring correct application of integration rules in the context of the differential equation.

maiad
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Homework Statement


Suppose that y(x) is the solution to the initial problem, y'=y(1-x), y(1)=e
find y(2)

Homework Equations


The Attempt at a Solution


This is my initial attempt:
[itex]\frac{dy}{dx}[/itex]=y(1-x)
[itex]\frac{dy}{y}[/itex]=(1-x)dx

i then integrated both side to get:
lny=-ln(1-x)+C

and here's the problem, if i plug in 1 to find c, ln(1-x) does not exist at that point...
 
Last edited:
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I tried another approach:
y'+yx=y
(yx)'=y

integrated both sides:
yx=[itex]\frac{1}{2}[/itex]([itex]y^{2}[/itex]) +C

but with this approach i wasn't able to solve for y
 
maiad said:
[itex]\frac{dy}{y}[/itex]=(1-x)dx

i then integrated both side to get:
lny=-ln(1-x)+C
First of all, you integrated the right hand side incorrectly. Remember, you're just integrating a polynomial.
 
maiad said:
I tried another approach:
y'+yx=y
(yx)'=y
Wait a minute, (yx)'=xy'+yx'=xy'+y, not y'+yx.
 

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