Separable Differential Equation

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Homework Help Overview

The discussion revolves around solving a separable differential equation given by y' = (x)/(1+2y) with the initial condition y(-1) = 0. Participants are exploring the steps involved in manipulating the equation and the implications of their algebraic operations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of multiplying both sides of the equation by (1+2y) and the subsequent algebraic manipulation. There is confusion regarding the validity of subtracting 1 from both sides of the equation and its impact on the terms involved.

Discussion Status

The discussion is active, with participants questioning the correctness of their algebraic steps. Some guidance is offered regarding the implications of operations on both sides of the equation, but there is no consensus on the resolution of the confusion.

Contextual Notes

Participants are grappling with the rules of algebra as they apply to differential equations, particularly in the context of manipulating terms that are multiplied by derivatives. There is a focus on understanding the role of parentheses in expressions.

MathWarrior
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y' = (x)/(1+2y)

y(-1) = 0

trying to find the answer I do the following:

multiply both sides by (1+2y)
(1+2y) * dy/dx = x

i subtract 1 from both sides.. but for some reason this is wrong? why?

2y * dy/dx = x - 1

2y dy = x-1 dx

integrate..

y^2 = (x^2-x+C)/2

y = sqrt((x^2-x+C)/2) )

this however is wrong I was not suppose to subtract the 1 from both sides.. what exactly is wrong with doing that?
 
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MathWarrior said:
y' = (x)/(1+2y)

y(-1) = 0

trying to find the answer I do the following:

multiply both sides by (1+2y)
(1+2y) * dy/dx = x

i subtract 1 from both sides.. but for some reason this is wrong? why?
If you subtract 1 from both sides it will be (1+2y) * dy/dx-1 = x-1.

(1+2y) * dy/dx = x is not the same as 2y * dy/dx = x - 1 as both 1 and 2y is multiplied by dy/dx in the original equation.

ehild
 


ehild said:
If you subtract 1 from both sides it will be (1+2y) * dy/dx-1 = x-1

even if its multiplied by it like that is?

I mean I am seeing it in my mind sort of like

1+5a = b
in which I would think to solve such an algebraic equation it would be more like

-1 on both sides
5a=b-1 ?
 


Do you know what are parentheses for? I just wonder how you subtract 2 from (2+5)10.

ehild
 

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