Science:Math Exam Resources/Courses/MATH101/April 2014/Question 06/Solution 1

By the separation of variables,

${\begin{aligned}x{\frac {dy}{dx}}&=y^{2}-y\\{\frac {dy}{y^{2}-y}}&={\frac {dx}{x}}\\\int {\frac {1}{y^{2}-y}}dy&=\int {\frac {1}{x}}dx\\\int {\frac {1}{y(y-1)}}dy&=\ln |x|+C\end{aligned}}$

For the $y$ integral, we can decompose

${\begin{aligned}{\frac {1}{y(y-1)}}={\frac {1}{y-1}}-{\frac {1}{y}},\end{aligned}}$

and therefore, we have,

${\begin{aligned}\int {\frac {1}{y-1}}-{\frac {1}{y}}dy&=\ln |x|+C\\\ln |y-1|-\ln |y|&=\ln |x|+C\\\ln \left|{\frac {y-1}{y}}\right|&=\ln |x|+C.\end{aligned}}$

This implies

${\begin{aligned}\left|{\frac {y-1}{y}}\right|=e^{C}|x|.\end{aligned}}$

The initial condition is $y(1)=-1$, and so we get $2=e^{C}$. Thus,

${\begin{aligned}\left|{\frac {y-1}{y}}\right|=2|x|\end{aligned}}.$

Furthermore, when we use the initial data both quantities are positive and so we can drop the absolute value signs:

${\begin{aligned}{\frac {y-1}{y}}&=2x\\1-{\frac {1}{y}}&=2x\\{\frac {1}{y}}&=1-2x\\y&={\frac {1}{1-2x}}.\end{aligned}}$