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I'm having difficulty understanding equilibrium solutions. Consider trying to find the equilibrium solution(s) of the following ODE:

\begin{equation} y'=x^2-y^4 \end{equation}

Firstly, I set $y'=0$ and solved for y:

\begin{align} 0&=x^2-y^4 \\ y^4&=x^2 \\ y&=\pm\sqrt{x} \end{align}

Now my lecturer/workbook specify that there is only an equilibrium solution when $y=c$, for some constant $c$. This would mean that solutions for $y$ that are functions of another variable are not equilibrium solutions. What then is the meaning of this solution? Just that it isn't one?

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    You *assume* $y'(x) = 0$ and then derive that $y(x) = \pm \sqrt{x}$. However if $y(x) = \sqrt{x}$ (as a function) then $y'(x) \not = 0$ so it does not satisfy your assumption that it is an equilibrium solution. – 2017-01-05

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What you found out is not a solution. Just try to substitute in the equation.

By definition, an equilibrium point is a constant solution of the equation. In this case, substituting $y=c$ leads to $x^2=c^2$ and so there are no solutions (note that by definition a solution is a $C^1$ function defined on some open interval, not on a point).

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    Thanks for the answer, but what do you mean by $C^1$ (and this definition of a solution?) is it explaining why the equilibrium solutions are horizontal lines and not just a point? – 2017-01-05
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    Solutions are functions, in this case of the variable $x$. They should be $C^1$ with respect to this variable, on some open set. – 2017-01-05
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As John.B has explained, this equation does not have an equilibrium solution. What you have found is the isocline curve corresponding to slope 0. That is, all points along the curve $y(x)=\pm\sqrt{x}$ will have zero slope in the slope field of the differential equation.

If an isocline curve corresponding to slope zero, happens to be a horizontal line(i.e. if it happens to be a solution to the differential equation), then that isocline curve is an equilibrium solution.

P.S- Isocline curves are solutions to the equation: $ y'(x) =f(x,y)= c $