# Trouble with a separable differential equation

1. Aug 2, 2014

### eric_999

I have this equation: dy/dx = 1-y^2, so then dy/(1-y^2) = dx, so ∫dy/(1-y^2) = dx --->
∫(A/(1+y) + B/(1-y))dy = x + C. I rewrite it again: (A - Ay + B + By)/(1+y)(1-y) = 1/(1+y)(1-y) so I get A+B = 1, and B-A = 0, so B = A, and therefore 2A = 1. so A & B = 1/2.

So 1/2∫(dy/(1+y) + 1/2∫dy/(1-y) = x + C ---> 1/2(ln|1+y| -ln|1-y|) = x + C ---> ln(|1+y|/|1-y|) = 2x + 2c = 2x + C, since C is arbitrary. At last I get |1+y|/|1-y| = e^(2x + c) ?=? |(1+y)/(1-y)| = e^(2x + C). How do I solve for y? Do Ihave to split it up into two separate cases, one when (1+y)/1-y) = e^(2x +C) and one where -(1+y)/(1-y) = e^(2x + C), or is this whole method simply just wrong?

2. Aug 2, 2014

### pasmith

You have $$\left|\frac{1 + y}{1 - y}\right| = Ae^{2x}.$$ You will need to deal with the cases $y < -1$, $-1 < y < 1$ and $y > 1$ separately.

3. Aug 2, 2014

### eric_999

Hmm, so if I have |(1+y)/(1-y)|, and I consider case 1, where -1<y<1 I get that |1+y| > 0 and |1-y|> 0, so the whole expression is positive, so then A must be > 0. So then I guess I have (1+y) = (1-y)Ae^2x, so y + yAe^2x = Ae^2x - 1, so y(1 +Ae^2x) = Ae^2x -1 so y = (Ae^2x -1)/(Ae^2x + 1).

When y<-1, (1+y) < 0, so I can write it like|1+y| = -(1+y). (1-y) on the other hand is positive, so |1-y| = (1-y), so then i have -(1+y) = (1-y)Ae^2x, so 1+y = -(Ae^2x - yAe^2x) ---> y = yAe^2x -Ae^2x -1, so y(1-Ae^2x) = -Ae^2x -1 ---> y = -(Ae^2x +1)/-(Ae^2x -1) = (Ae^2x +1)/Ae^2x -1). I dunno, in the book, they say im gonna get y = +1, y = -1, and y = (Ae^2x -1)/(Ae^2x + 1). I can't see how they get y = +-1!

4. Aug 2, 2014

### eric_999

Hmm, so if I have |(1+y)/(1-y)|, and I consider case 1, where -1<y<1 I get that |1+y| > 0 and |1-y|> 0, so the whole expression is positive, so then A must be > 0. So then I guess I have (1+y) = (1-y)Ae^2x, so y + yAe^2x = Ae^2x - 1, so y(1 +Ae^2x) = Ae^2x -1 so y = (Ae^2x -1)/(Ae^2x + 1).

When y<-1, (1+y) < 0, so I can write it like|1+y| = -(1+y). (1-y) on the other hand is positive, so |1-y| = (1-y), so then i have -(1+y) = (1-y)Ae^2x, so 1+y = -(Ae^2x - yAe^2x) ---> y = yAe^2x -Ae^2x -1, so y(1-Ae^2x) = -Ae^2x -1 ---> y = -(Ae^2x +1)/-(Ae^2x -1) = (Ae^2x +1)/Ae^2x -1). I dunno, in the book, they say im gonna get y = +1, y = -1, and y = (Ae^2x -1)/(Ae^2x + 1). I can't see how they get y = +-1!

5. Aug 4, 2014

### pasmith

$dy/dx = 0$ when $y = \pm 1$.