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)