How to Solve the Differential Equation (x^2 - 1)y' + 2xy = x?

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Discussion Overview

The discussion revolves around solving the differential equation (x^2 - 1)y' + 2xy = x, focusing on methods for finding solutions and integrating factors. Participants engage in various approaches to rearranging the equation and solving it, with an emphasis on showing full working steps.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant requests a full working solution to the differential equation.
  • Another suggests rearranging the equation to find an integrating factor, proposing that it should lead to an exact differential.
  • A different participant mentions rewriting the equation in standard form and identifies P as 2x for the integrating factor calculation.
  • One participant claims to have solved the equation, presenting a solution without an arbitrary constant.
  • Another participant corrects the previous solution by noting the need for an arbitrary constant in the final expression.
  • A later reply expresses gratitude for the assistance received in the discussion.
  • One participant provides a detailed solution, showing the integration process and the final form of y in terms of x.

Areas of Agreement / Disagreement

There are multiple competing views regarding the methods for solving the differential equation, and the discussion remains unresolved as participants present different approaches and solutions without reaching a consensus.

Contextual Notes

Some participants' solutions depend on specific assumptions about the integrating factor and the form of the equation, which may not be universally agreed upon. The presence of arbitrary constants in solutions is also a point of contention.

josek6
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Solve showing full working
(x^2 -1)y' + 2xy = x

You must show full working.
 
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No full workings here, were fresh out.
Try to rearrange and find an integrating factor u such that
u(x^2 -1)y' + u(2xy - x)

is and exact differential
 
Find the integerating factor "u"
So I have to rewrite the equation in the standard form y' + Py =Q
IF = e ^ ∫P.dx
where P = 2x ?
 
Thanks...I solved it my solution is
y(x^2 -1)=(x^2)/2
 
Good, but you need the arbitrary constant
y(x^2 -1)=(x^2)/2+C
 
I must commend you guys for all the assistance and I will surely recommend you guys to my friends. Thanks again.
 
(x^2 - 1 ) y' + 2xy = ((x^2-1)y)' = x;-->
(x^2-1)y = \int{x};-->
(x^2-1)y = 0.5 x^2 + C

y = (0.5 x^2 + C) / ( x^2 - 1 )
 

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