Why Equilibria Need Special Attention
Separating variables often requires dividing by an expression involving the dependent variable \(y\). If that expression can equal zero, division restricts the calculation to places where it is nonzero. A solution that makes the expression zero for all time may therefore disappear from the separated equation.
The missing solution is often a constant equilibrium solution: the dependent variable stays at a value where the differential equation’s right-hand side is zero. Finding such values before dividing is an important part of solving a separable differential equation correctly.
What Is an Equilibrium Solution?
For an autonomous differential equation \(y'=f(y)\), a constant function \(y(t)=c\) is a solution exactly when \(f(c)=0\). Its derivative is zero, so the equation requires the right-hand side to be zero at that value.
The same check applies when the equation has a time-dependent coefficient, such as \(y'=a(t)h(y)\). If \(h(c)=0\), then \(a(t)h(c)=0\) for every \(t\), so \(y(t)=c\) is a solution. Conversely, if \(a(t)\) is not identically zero on the time interval, a constant solution must have \(h(c)=0\).
This follows directly from the definition of a solution. For \(y(t)=c\), we have \(y'(t)=0\). Substituting into \(y'=a(t)h(y)\) gives
for every \(t\) in the interval. If \(h(c)=0\), this equality holds for all \(t\). If there is a time \(t\) for which \(a(t)\ne 0\), the equality at that time requires \(h(c)=0\). This establishes the criterion.
Find Equilibria Before Separating
Suppose the equation is written in factored form as \(y'=a(t)h(y)\). The zeros of \(h(y)\) are the first values to check. If the right-hand side has several factors depending on \(y\), set each such factor equal to zero and test the resulting values in the original equation.
Worked Example: Finding Equilibrium Values from Factors
Find the constant equilibrium solutions of
For a constant solution \(y(t)=c\), the derivative is \(y'=0\). The equation therefore requires
A product is zero when at least one factor is zero:
Thus the candidate equilibrium solutions are \(y(t)=2\) and \(y(t)=-1\). Check each in the original equation. For \(y(t)=2\), the derivative is zero and the right-hand side is \((2-2)(2+1)=0\). For \(y(t)=-1\), the derivative is zero and the right-hand side is \((-1-2)(-1+1)=0\). Both are solutions.
Why Division Can Lose Solutions
To separate an equation \(y'=a(t)h(y)\), one may divide by \(h(y)\), giving
This step is valid only at points where \(h(y)\ne 0\). If \(h(c)=0\), the constant function \(y(t)=c\) makes the denominator zero, so it cannot appear in the divided equation. The division has not shown that the constant function fails to solve the original equation; it has simply excluded it from the calculation.
On an interval where \(h(y)\ne 0\), separation and integration can produce nonconstant solutions. Those solutions do not automatically include the excluded equilibria. The complete solution set must account for both: the solutions obtained on the nonzero branches and any equilibrium solutions found from the original equation.
Write the equation in a form that makes its \(y\)-dependent factors visible.
Set the \(y\)-dependent factor or factors equal to zero and verify the resulting constants in the original equation.
State that the divided equation applies where the divisor is nonzero.
Include the valid constant solutions alongside the solutions obtained by separation.
Worked Separation with a Time-Dependent Factor
A coefficient depending on time does not change the need to check zeros of the \(y\)-dependent factor. It can, however, be useful to see explicitly how a constant solution is excluded by division.
Worked Example: Separating an Equation That Has an Equilibrium
Solve \(y'=t(y-3)\) for solutions, and find the solution satisfying \(y(0)=5\).
First, check for a constant equilibrium. For \(y(t)=3\), the derivative is zero and the right-hand side is \(t(3-3)=0\) for every \(t\). Thus \(y(t)=3\) is an equilibrium solution.
For other solutions, work where \(y\ne 3\). Divide by \(y-3\) and separate variables:
Integrating both sides gives
Exponentiating gives \(|y-3|=e^C e^{t^2/2}\). Allowing either sign and combining the constant factors into one real constant \(K\ne 0\) gives the non-equilibrium solutions
Apply \(y(0)=5\). Since \(e^0=1\),
The particular solution is \(y=3+2e^{t^2/2}\). Check the initial condition: \(y(0)=3+2=5\). Its derivative is \(y'=2e^{t^2/2}\cdot t=t(2e^{t^2/2})\), while the right-hand side is \(t(y-3)=t(2e^{t^2/2})\); the two sides agree.
The separated calculation required \(y\ne 3\), so it did not derive the equilibrium \(y=3\). The formula with \(K=0\) would represent that equilibrium, but setting \(K=0\) is an extension of the resulting expression, not a value obtained through the division step.
A Zero Factor Can Give a Solution Missed by Separation
Worked Example: The Equilibrium of \(y'=y^2\)
Find the equilibrium solution and solve the equation for the initial value \(y(0)=1\).
For a constant \(y(t)=c\), the equation requires \(0=c^2\), so \(c=0\). Check in the original equation: the derivative of \(y(t)=0\) is zero and the right-hand side is \(0^2=0\). Thus \(y(t)=0\) is an equilibrium solution.
For solutions with \(y\ne 0\), separate and integrate:
Use \(y(0)=1\) to determine \(C\):
Then \(-1/y=t-1\), so \(y=1/(1-t)\), on any interval containing \(0\) that does not include \(t=1\). Check the initial value: \(y(0)=1\). Differentiate and compare with the right-hand side:
The differential equation is satisfied. The equilibrium \(y=0\) is not obtained by dividing by \(y^2\), because that division assumes \(y\ne 0\).
Do Not Turn an Excluded Case into an Invalid One
A common mistake is to divide by a factor and then treat the resulting equation as though it includes every solution of the original equation. For \(y'=y(y-1)\), for example, dividing by \(y(y-1)\) gives
That equation applies only where \(y\ne 0\) and \(y\ne 1\). The original equation has two constant solutions, \(y=0\) and \(y=1\), because its right-hand side is zero at both values. The divided equation cannot represent either one.
Correction: Division by \(y\) is valid only on a part of a solution where \(y\ne 0\). It says nothing by itself about whether \(y=0\) solves the original equation. Substitute \(y=0\) into that original equation to decide.
The safe reasoning is not to assume that every solution avoids a zero factor. Instead, test the zero values in the original equation first. Then keep the separated calculation within the domain where its divisions are valid.
Why the Distinction Matters
An equilibrium is a genuine solution, not a technical exception to discard. It can also affect how an initial condition is handled. If the initial value is an equilibrium value, dividing by the factor that vanishes there is not a valid first step for that initial-value problem. Check the constant function directly in the original equation.
There is also a useful distinction between a solution formula obtained after division and the full set of solutions. The former describes solutions on domains where the divisor is nonzero. The latter must also include any valid equilibria excluded by that restriction. Stating where division is valid makes this distinction precise.
Check Your Understanding
For each question, keep the original differential equation in view when deciding whether division is valid.
- Find the constant equilibrium solutions of \(y'=(y-4)(y+2)\).
- For \(y'=t(y+1)\), why does dividing by \(y+1\) not include every solution?
- For \(y'=y^3\), find the constant equilibrium and state the restriction required when separating variables.
- A student divides an equation by \(y-2\) and concludes that \(y=2\) cannot be a solution. What is the flaw in this conclusion?
- What should you check before dividing by a factor that depends on \(y\)?
