Solving a Simple Differential Equation: x.dy/dx+1-y^2=0

[CNS92]

Homework Statement

Solve x.dy/dx+1-y^2=0

Homework Equations

The Attempt at a Solution

Separate:
dy/(y^2-1) = dx/x

The LHS can be broken into
dy/(2(y-1))-dy/(2(y+1))

Integrating:

Log[y-1]/[y+1] = log[x^2] + c

Given x=1 when y=0 c=0

Y-1=x^2.y+x^2

Y=(x^2+1)/(x^2-1)

The answer is the reciprocal of mine, can anyone see my mistake?

Thanks

 

[CompuChip]

Actually it's not the reciprocal, it is the negative:
y = (x^2+1)/(1-x^2) = -(x^2+1)/(x^2-1)
is the correct answer.

There is a simple algebraic error in the last step, everything is correct up to
y-1=x^2.y+x^2

Then,
y - x^2 y = 1 + x^2
(1 - x^2) y = (1 + x^2)
y = (1 + x^2) / (1 - x^2)

 

[CNS92]

http://www.wolframalpha.com/input/?i=solve+x*dy/dx+++(1-y^2)=0+

Weirdly, I can't get this question right doing it as above, but if I divide by (1-y^2) it does work.

If you drop modulus signs, the constant wasn't 0 but 0.5log(-1).

Not really sure why the mod signs mess it up like that..

 

[lurflurf]

The answer is
y=(1-x^2)/(1+x^2)
I think you lost a sign with the logarithms.
y=(C-x^2)/(C+x^2)
y=1
for example are solutions
Given x=1 when y=0 we see C=1.
Your answer is the case C=-1 which does not satisfy the initial conditions, but since you were working in absolute value the condition |C|=1 was satisfied.