Polynomial Factors in Separable Equations
A polynomial in the dependent variable often makes a differential equation separable. Its roots are especially important: when the dependent variable stays at one of those roots, the derivative can be zero, producing a constant solution. Any separation method must account for these equilibria before dividing by a polynomial expression.
For a nonconstant solution, separation leads to an integral involving the reciprocal of that polynomial. Some such integrals are immediate, while others require methods that will be treated separately. Here we focus on polynomial factors that can be integrated directly, especially the logistic factor \(y(1-y)\).
The equation \(y'=g(x)P(y)\) can be separated wherever \(P(y)\ne 0\): \(\frac{dy}{P(y)}=g(x)\,dx\). The restriction matters. At a root \(r\) of \(P\), division by \(P(y)\) is not allowed, so the constant solution \(y=r\) must be recorded separately.
Equilibria and the Nonconstant Solutions
For an equation \(y'=g(x)P(y)\), every real root \(r\) of \(P\) gives an equilibrium solution \(y(x)=r\). This follows by substitution: its derivative is zero, and its right-hand side is \(g(x)P(r)=0\). On any interval where a solution has \(P(y(x))\ne 0\), the equation can instead be divided by \(P(y)\) and integrated.
This gives a useful overall plan. First find the roots of \(P\) and list the corresponding constant solutions. Then solve the separated equation only on intervals where its denominator is nonzero. Finally, use the initial condition to select the solution. For a polynomial \(P\), the reciprocal \(1/P(y)\) is a rational function; not every such integral should be expected to have the same simple form. The examples below use factors whose integrals can be obtained directly.
Set \(P(y)=0\). Each real root gives a constant equilibrium solution.
For a non-equilibrium solution, divide by \(P(y)\) only on an interval where \(P(y)\ne 0\).
Determine the integration constant and restrict the formula to an interval on which it is defined.
Differentiate the proposed solution and substitute it into the original equation, including its initial value.
Worked Example: A Logistic Initial-Value Problem
Worked Example: A Logistic Initial-Value Problem
Solve \(y'=y(1-y)\) subject to \(y(0)=\frac13\).
First, find the roots of the polynomial factor \(P(y)=y(1-y)\). They are \(y=0\) and \(y=1\), so the equation has the equilibrium solutions \(y(x)=0\) and \(y(x)=1\). The initial value is neither root, so we may separate variables along the solution.
For \(y\ne 0,1\), divide by \(y(1-y)\). The identity \(\frac{1}{y(1-y)}=\frac1y+\frac{1}{1-y}\) holds because the right-hand side has numerator \(1-y+y=1\) over denominator \(y(1-y)\). Thus
Integrating gives
Exponentiating and absorbing the constant sign into a nonzero constant \(A\), we obtain \(\frac{y}{1-y}=Ae^x\). Solving for \(y\) gives \(y=\frac{Ae^x}{1+Ae^x}=\frac{1}{1+Ce^{-x}}\), where \(C=1/A\).
Apply \(y(0)=\frac13\):
Therefore the solution is \(y(x)=\frac{1}{1+2e^{-x}}\). Its denominator is positive for every real \(x\), so this formula is defined for all real \(x\).
To check it, differentiate:
Meanwhile,
Thus \(y'=y(1-y)\). At \(x=0\), \(y(0)=1/(1+2)=1/3\), as required.
Why the Logistic Formula Works
The logistic equation illustrates why separation produces a logarithm and why the resulting formula has a denominator. The key step is the identity used in the example; it can be verified by combining the fractions. The following derivation gathers the steps into a general solution for non-equilibrium initial values.
Integrating gives \(\ln|y/(1-y)|=x+C_0\), where \(C_0\) is a real constant. Exponentiating shows that \(y/(1-y)=Ae^x\) for a nonzero real constant \(A\); the sign of \(A\) accounts for the absolute value. Rearranging this equation yields \(y=Ae^x/(1+Ae^x)\). Writing \(C=1/A\) gives the equivalent form below, on intervals where its denominator is nonzero:
Every formula in this family satisfies the equation wherever it is defined. In fact, writing \(D=1+Ce^{-x}\) gives \(y=D^{-1}\), so \(y'=Ce^{-x}/D^2\). Also \(1-y=Ce^{-x}/D\), and hence \(y(1-y)=Ce^{-x}/D^2=y'\). The constant solutions \(y=0\) and \(y=1\) complete the list. The formula with \(C\ne 0\) does not represent \(y=0\); the equilibrium must remain on the list separately.
If an initial value is \(y(x_0)=y_0\), with \(y_0\ne 0,1\), substitution into the formula gives
This determines the member of the nonconstant family that passes through the initial point. If \(y_0=0\) or \(y_0=1\), the corresponding equilibrium solution satisfies the initial condition. Since the right-hand side \(y(1-y)\) is a polynomial, the uniqueness result for smooth differential equations ensures a solution starting at either equilibrium remains there on its interval of definition.
Worked Example: A Polynomial Square
Worked Example: A Polynomial Square
Solve \(y'=x(1-y)^2\) with \(y(0)=0\).
The polynomial factor \((1-y)^2\) vanishes at \(y=1\), so \(y(x)=1\) is an equilibrium solution. The initial value is \(0\), not \(1\), so separate on an interval where \(1-y\ne0\):
The derivative of \(1/(1-y)\) with respect to \(y\) is \(1/(1-y)^2\). Therefore integration gives
Using \(y(0)=0\) gives \(1=C\), so \(\frac{1}{1-y}=1+\frac{x^2}{2}\). Solving for \(y\),
The denominator \(2+x^2\) is positive for every real \(x\). To check the differential equation, differentiate and simplify:
Also \(1-y=2/(2+x^2)\), so \(x(1-y)^2=4x/(2+x^2)^2=y'\). Finally, \(y(0)=0/(2+0)=0\). Both the equation and the initial condition are satisfied.
Worked Example: A Repeated Equilibrium Factor
Worked Example: A Repeated Equilibrium Factor
Solve \(y'=xy^2\) with \(y(0)=1\).
The polynomial factor \(y^2\) has the root \(0\), so \(y=0\) is an equilibrium solution. The given initial value is nonzero, so divide by \(y^2\) on the interval of the nonconstant solution:
Since \(\int y^{-2}\,dy=-y^{-1}\), integration gives
At \(x=0\), \(y=1\), so \(-1=C\). Thus \(-1/y=x^2/2-1\), or \(1/y=1-x^2/2\). Solving explicitly,
The denominator vanishes at \(x=\pm\sqrt2\). The interval containing the initial time \(0\) on which the formula is defined is \((-\sqrt2,\sqrt2)\). To check, differentiate:
Since \(y^2=1/(1-x^2/2)^2\), the right-hand side \(xy^2\) equals \(x/(1-x^2/2)^2=y'\). Also \(y(0)=1\). The formula therefore satisfies the initial-value problem throughout its interval of definition.
A Common Pitfall: Dividing Away an Equilibrium
A separation calculation is valid only where the expression used as a divisor is nonzero. Consider the tempting calculation for \(y'=y(1-y)\): divide by \(y(1-y)\), then solve the resulting logarithmic equation. This division assumes \(y\ne0\) and \(y\ne1\). It cannot produce either constant equilibrium.
The correction is to substitute the roots into the original equation before dividing. The constant functions \(y=0\) and \(y=1\) both satisfy it. Then solve the separated equation on the non-equilibrium branches. In general, always distinguish solutions at roots of \(P\) from solutions obtained by dividing by \(P(y)\).
What Polynomial Factors Tell You
The roots of the polynomial factor give more than a list of solutions that must be checked. They divide the \(y\)-axis into intervals on which the sign of \(P(y)\) does not change. For an autonomous equation \(y'=P(y)\), that sign indicates whether a solution is increasing or decreasing while it remains in one of those intervals. The earlier material on stability of equilibria uses these sign changes to describe nearby behavior.
For an equation with a coefficient \(g(x)\), the sign of \(g(x)\) also affects the direction of change, so the polynomial alone does not determine it at every time. In either case, roots remain essential: they identify constant solutions and mark the values at which a division used in separation would fail.
For an initial-value problem, a practical checklist is:
- Find every real root of the polynomial factor and write down the associated equilibrium solutions.
- If the initial value is a root, use the corresponding constant solution.
- Otherwise, separate variables only where the polynomial factor is nonzero.
- Apply the initial condition before deciding which interval of the formula is relevant.
- Differentiate and substitute to check the differential equation and verify the initial value.
These steps keep the algebra connected to the original differential equation. They also prevent two common errors: losing an equilibrium by division and accepting an explicit formula at points where its denominator vanishes.
Check Your Understanding
For each question, keep track of where division is permitted and which solutions must be considered separately.
- For \(y'=y(1-y)\), what are the equilibrium solutions, and why are they not obtained by dividing by \(y(1-y)\)?
- For \(y'=x(1-y)^2\), identify the equilibrium and state the condition needed before dividing by \((1-y)^2\).
- Separate and integrate \(y'=y^2\) on an interval where \(y\ne0\). What equilibrium must be listed separately?
- For the logistic formula \(y=1/(1+Ce^{-x})\), substitute it into \(y'=y(1-y)\) to verify the equation.
- For \(y'=xy^2,\ y(0)=1\), where does the explicit formula's denominator vanish, and which interval contains the initial time?
