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)