Science:Math Exam Resources/Courses/MATH103/April 2011/Question 06 (a)/Solution 1

The steady state solutions are solutions to the differential equation that are constant. That is, $y(x)={\rm {constant}}$. For this to be possible, the derivative of $y$ has to be zero.

Therefore, to be consistent with the differential equation, we have that

$0=(y^{2}-1)x$.

For this equation to be true, we require that $y=\pm 1$ or $x=0$. Since $x=0$ does not fit the definition of a steady state solution, we are left with the steady solutions $y=\pm 1$.