# 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