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

##### 1 Answer

Apr 28, 2018

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

#### Explanation:

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