Question

Solve separable differential equation:

Answer 1

http://i39.tinypic.com/j0g1e1.jpg

Answer 2

You have that \[\frac{dy}{dx}=xy^2-x-y^2+1=x(y^2-1)-(y^2-1)=(x-1)(y^2-1)\] \[\frac{dy}{y^2-1}=(x-1)dx\] Using partial fractions you can rewrite the left hand side as \[\frac12\left(\frac{1}{y-1}-\frac{1}{y+1}\right)dy=(x-1)dx\] Integrating both sides then gives \[\frac12(\ln|y-1|-\ln|y+1|)=\frac12x^2-x+C\] Multiplying both sides by 2 and combining the logs gives \[\ln\left|\frac{y-1}{y+1}\right|=\ln\left|1-\frac{2}{y+1}\right|=x^2-2x+C\] Exponentiating each side gives \[1-\frac{2}{y+1}=Ae^{x^2-2x}\] And then simplifying gives yours final answer of \[y=\frac{2}{1-Ae^{x^2-2x}}-1\]

Answer 3

okay, I got stuck at the very first line with the factoring. Thank you so much.