Differential Eq. Last Step Solution Separating Variables

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Discussion Overview

The discussion revolves around solving a differential equation using the method of separation of variables. Participants are focused on the final steps of deriving an explicit solution for the equation and are seeking clarification on how to manipulate the resulting expression.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the differential equation and expresses difficulty in deriving the explicit solution from the equation y^2 + 3y = x^3 - x + C.
  • Another participant suggests completing the square on the left-hand side and taking the square root to solve for y.
  • Multiple participants reiterate the quadratic form y^2 + 3y + (x - C) = 0 and recommend using the quadratic formula to find y.
  • There is a question regarding the omission of the x^3 term in a previous response, with one participant suggesting it was inadvertently left out.
  • One participant humorously claims to have removed the x^3 term through "carelessness," while others note that the methods discussed are mathematically equivalent.
  • It is mentioned that if there is an initial condition, it may influence the choice between the two possible solutions derived from the quadratic formula.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the quadratic equation, but there is some confusion regarding the treatment of the x^3 term, indicating a lack of consensus on that specific point.

Contextual Notes

The discussion does not resolve the question of how to handle the x^3 term, and participants express varying degrees of clarity on the steps involved in solving the quadratic equation.

knowLittle
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Solve each differential equation. Express the general solution in explicit form.

y' = (3x^2 -1) / (3+2y)

So, I will skip many steps, because they are easy. However, I am stuck in one of the last ones.
y^2 +3y = x^3- x +C

y(y+3)= x^3 - x +C

I have seen the solution for y, but I don't understand how it is derived. Can someone help?
 
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Try completing the square on the left hand side, and then taking the square root of both sides of the equation.
 
y(y+ 3)= y^2+ 3y= x^3- x+ C

y^2+ 3y+ (x- C)= 0.

Solve that quadratic equation using the quadratic formula.
 
HallsofIvy said:
y(y+ 3)= y^2+ 3y= x^3- x+ C

y^2+ 3y+ (x- C)= 0.

Solve that quadratic equation using the quadratic formula.

Hello HallsofIvy, How did you get rid of x^3 ?
 
knowLittle said:
Hello HallsofIvy, How did you get rid of x^3 ?

He appears to inadvertently left it out. The method HallsofIvy suggested and the method I suggested are mathematically equivalent.

Chet
 
knowLittle said:
Hello HallsofIvy, How did you get rid of x^3 ?

Chestermiller said:
He appears to inadvertently left it out.
Yes.
Chestermiller said:
The method HallsofIvy suggested and the method I suggested are mathematically equivalent.
Move all of the terms to the left side, and you'll have a quadratic in y. It's probably simpler at this point to just use the Quadratic Formula to solve for y, which will have two parts separated by ±, as is usually the case.

If there is an initial condition, it might lead you to choose one or the other of the two values.
 
knowLittle said:
Hello HallsofIvy, How did you get rid of x^3 ?
I waved my magic wand and uttered a spell of "Carelessness"!
 

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