Question

Find the solution of the given initial value problem in explicit form
y'=(1-2x)/y ; y(1)=-2

Answer 1

\[\frac{dy}{dx}y =1-2x\\ydy=(1-2x)dx\]
integral both sides I got the answer is y =+-sqrt(2x-2x^2 +c
solve for c , I have c=4
so the answer should be +- sqrt(2x+2x^2+4
However, the answer in book reject positive answer, why?

Answer 2

are you saying the answer is
\[ y= \pm \sqrt{2x - 2x^2 +4} \]

at x=0 you would get
\[ y = \pm\sqrt{4} = \pm 2\]
based on y(1)= -2, you can only use the - sign solution