# Thread: compute the general solution (1-x)y'=y^2

1. ## compute the general solution (1-x)y'=y^2

compute the general solution (1-x)y'=y^2

Can anyone, solve this step by step?

2. ## Re: compute the general solution (1-x)y'=y^2

Where did you get this problem? From a course in Calculus or Differential Equations? In either case you certainly should know how to integrate:
[tex](1- x)y'= (1- x)\dfrac{dy}{dx}= y^2[tex]

$\displaystyle \dfrac{dy}{y^2}= y^{-2}dy= \frac{dx}{1- x}$
Can you integrate those?

3. ## Re: compute the general solution (1-x)y'=y^2

dy/(y^2)=dx/(1-x)
y^-2dy=ln (1-x)+c
-1/y=ln (1-x)+c

I stock from here.

4. ## Re: compute the general solution (1-x)y'=y^2

It is ok just solve for y.......(make y the subject)

5. ## Re: compute the general solution (1-x)y'=y^2

One thing you can do is "flip" both sides:
$\displaystyle \frac{-1}{y}= ln|1- x|+ c$ becomes
$\displaystyle -y= \frac{1}{ln|1- x|+ c}$

(Notice the absolute value. ln(x) isn't defined for $\displaystyle x\le 0$. If you know x< 1, for example, if you given a value for y(0)= 0, you can use just "1- x". On the other hand, if you have reason to believe x>1, say you are given a value of y(1), you need "x- 1".)