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))