# Solve : ( y^2 - 4 )dx - dy = 0

1. Apr 1, 2005

### Naeem

I did the following:

dy/(y^2 -4 ) = dx

Integral ( -1/4 /y + 2 + 1/4 / ( y-2) . dy = Integral dx

-1/4 ln | y + 2 | + 1 /4 ln |y -2 | + C = x

Multiplying all thru by - 4 we get

ln | y + 2 | + ln | y - 2 | + 4C = 4x ( Note: 4C is another 'C'which is a bigger C)

ln | y + 2 / y - 2| + C = 4x

C = 4x - ln | y + 2 / y -2 |

If this is correct, can we simplify the solution any further.

2. Apr 1, 2005

### whozum

You ahve a lot of mismatched parentheses but,

$$\frac{1}{y^2-4} = \frac{1}{(y-2)(y+2)}$$

You cant separate these two in the denominator.

I would do this using the substitution y = 2sec(t)

3. Apr 1, 2005

### Naeem

Why not? Using Partial fractions you could

A/ something + B / something

4. Apr 1, 2005

### whozum

I dont think thats what he did, or if he did he didnt show the work. I can barely read his post.

5. Apr 2, 2005

### dextercioby

That "-4" multiplication is incorreclty made.You should have other signs...

Daniel.