What is the general solution of the differential equation #(y^2-1)xdx + (x+2)ydy = 0#?
1 Answer
Mar 11, 2018
the general (implicit) solution is:
# y^2 = 1+A(x+2)^4 e^(-2x) #
Explanation:
We have:
# (y^2-1)xdx + (x+2)ydy = 0 #
Which if we write in the form
# (x+2)ydy/dx = (1-y^2)x #
# :. y/(1-y^2) dy/dx = x/(x+2) #
Which is separable, so we can "separate the variables" to get:
# int \ y/(1-y^2) \ dy = int \ x/(x+2) \ dx #
Then we integrate (steps omitted) to get:
# -1/2ln|y^2-1| = x-2ln|x+2| + C #
# :. ln|y^2-1| = 4ln|x+2| -2x- 2C #
# :. ln|y^2-1| = ln(x+2)^4 -2x- 2C #
# :. y^2-1 = e^(ln(x+2)^4 -2x- 2C) #
# :. \ \ \ \ \ \ \ \ \ \ = e^(ln(x+2)^4) e^(-2x)e^(- 2C) #
# :. \ \ \ \ \ \ \ \ \ \ = A(x+2)^4 e^(-2x) #
Leading to the general (implicit) solution:
# y^2 = 1+A(x+2)^4 e^(-2x) #
