Ordinary differential equation:

[Edwardo_Elric]

Homework Statement

Obtain the general solution:
$$(1 - x)y' = y^2$$

Homework Equations

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

so my answer is:
$$-1 = y\ln{(c(1 - x))}$$

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?

 

[HallsofIvy]

Homework Statement

Obtain the general solution:
$$(1 - x)y' = y^2$$

Homework Equations

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

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

so my answer is:
$$-1 = y\ln{(c(1 - x))}$$

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?

 

[Edwardo_Elric]

oh yeah! forgot about that thanks a lot :D