Originally Posted by

**nivek0078** Thanks to all for your help! Please show me where i went wrong. I'm getting zeros as the answer for part a) and b)

Given 2xy(dy/dx) + y^2 - 1 = 0 find all solutions A) (1,0) B) (0,1) in the x-t plane.

Here's my workout:

2xy (dy/dx) +y^2 -1 = 0

(2xy/1-y^2) (dy/dx) =1

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

integrate both sides:

S (2y/(1-y^2) (dy/dx) = S 1/x dx

-lnly^2 - 1ldy = ln lxl + C

solve for y get rid of ln take e to both sides

e^(-lnly^2 - 1) = e^(ln lxl + C)

General solution:

y= square root x(e^(C/2))