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

• Oct 26th 2013, 07:53 AM
alden35
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?
• Oct 26th 2013, 08:32 AM
HallsofIvy
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]

$\dfrac{dy}{y^2}= y^{-2}dy= \frac{dx}{1- x}$
Can you integrate those?
• Oct 26th 2013, 08:47 AM
alden35
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.
• Oct 26th 2013, 09:14 AM
MINOANMAN
Re: compute the general solution (1-x)y'=y^2
It is ok just solve for y.......(make y the subject)
• Oct 26th 2013, 09:58 AM
HallsofIvy
Re: compute the general solution (1-x)y'=y^2
One thing you can do is "flip" both sides:
$\frac{-1}{y}= ln|1- x|+ c$ becomes
$-y= \frac{1}{ln|1- x|+ c}$

(Notice the absolute value. ln(x) isn't defined for $x\le 0$. 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".)