$y' = x/y - x/(1+y)$ ? by using separation of variable

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Answer 1 · 2018-04-28T01:54:52

$3y^2 + 2y^3 = 3x^2 + A$

We have:

$y' = x/y - x/(1+y)$

Which we can write as:

$y' = (x(1+y) - xy) / (y(1+y))$

$\ \ \ = (x+xy - xy) / (y(1+y))$

$\ \ \ = x / (y(1+y))$

$:. y + y^2 \ dy/dx = x$

Which is now a seperal DE, so we can "separate the variables" to get:

$int \ y + y^2 \ dy = int \ x \ dx$

Which we can now readily integrate:

$1/2y^2 + 1/3y^3 = 1/2x^2 + C$

$:. 3y^2 + 2y^3 = 3x^2 + A$

Which is the Implicit General Solution

y' = x/y - x/(1+y) y' = x/y - x/(1+y) ? by using separation of variable