Separable Differential Equation

  • #1
268
4
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?
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,510
1,897


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
 
  • #3
268
4


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 ?
 
  • #4
ehild
Homework Helper
15,510
1,897


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

ehild
 

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