# Ordinary differential equation:

1. Nov 14, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
Obtain the general solution:
$$(1 - x)y' = y^2$$

2. Relevant equations

3. The attempt at a solution
$$(1 - x)\frac{dy}{dx} = y^2$$

$$(1 - x)dy = y^2dx$$

$$\frac{dy}{y^2} = \frac{dx}{(1-x)}$$
integrating both sides:

i used ln on the constant at the right side
$$-\frac{1}{y} = \ln(1 - x) + \ln{c}$$

$$-1 = y\ln{(c(1 - x))}$$

the answer seems to be different at the back of my book w/c is
$$1 = y\ln{(c(1 - x))}$$ <<< no negative as before

can you tell me what's wrong with my soln?

2. Nov 14, 2007

### HallsofIvy

Staff Emeritus
Here is the problem. What do you get if you differentiate ln(1-x)? (Don't forget to use the chain rule!)

3. Nov 14, 2007

### Edwardo_Elric

oh yeah! forgot about that thanks alot :D