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

Can anyone, solve this step by step?

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- Oct 26th 2013, 06:53 AMalden35compute 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, 07:32 AMHallsofIvyRe: 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? - Oct 26th 2013, 07:47 AMalden35Re: 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, 08:14 AMMINOANMANRe: compute the general solution (1-x)y'=y^2
It is ok just solve for y.......(make y the subject)

- Oct 26th 2013, 08:58 AMHallsofIvyRe: 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".)