Why Partial Fractions Appear in Separable Equations
Some separable differential equations lead to integrals of rational functions of the dependent variable. A rational function is a quotient of polynomials. When its denominator factors into linear factors, partial fractions rewrite the quotient as a sum of simpler terms whose antiderivatives are known.
The algebra is only part of the method. A denominator zero may be a value where the differential equation is undefined, while a numerator zero may give an equilibrium solution. Dividing by a factor can also remove an equilibrium from the equation being integrated. The solution must therefore be checked against the original equation and its domain.
Partial Fractions for Linear Factors
For a proper rational function, the degree of the numerator is less than the degree of the denominator. If the denominator factors into distinct real linear factors, the function can be written as a sum with one constant numerator over each factor. If a linear factor is repeated, terms are needed for every power of that factor, up to its full multiplicity.
To find the constants, multiply both sides by the common denominator. The result is a polynomial identity, so the coefficients can be found by substituting convenient values or by matching coefficients. If the original rational function is not proper, first use polynomial division; the remainder can then be decomposed into partial fractions.
The basic antiderivatives used here are, for \(y\ne a\),
The absolute value in the logarithm matters: \(\ln|y-a|\) is defined on both intervals \(y<a\) and \(y>a\), though not at \(y=a\). Integration constants can differ on separate intervals separated by a denominator zero.
Why Separation and Integration Work
Consider an equation of the form \(y'=X(x)G(y)\). Suppose a solution stays in an interval of \(y\)-values where \(G\) is defined and nonzero. Let \(H\) be an antiderivative of \(1/G\) on that interval, so \(H'(y)=1/G(y)\). The chain rule gives a direct justification for separation:
Consequently, \(H(y(x))\) differs from an antiderivative of \(X(x)\) by a constant. Partial fractions are an algebraic tool for finding \(H\) when \(1/G(y)\) is rational. This argument requires that the solution remain where the equation is defined and where division by \(G(y)\) is valid. Any constant solution with \(G(y)=0\) must be checked separately in the original equation.
Worked Example: A Simple Rational Factor
Worked Example: A Simple Rational Factor
Solve \(y'=\frac{y-1}{y+2}\) for nonconstant solutions, and identify the equilibrium and the excluded value.
The equation is undefined at \(y=-2\). The numerator vanishes at \(y=1\), so \(y(x)=1\) is an equilibrium solution: substituting it into the original equation gives \(0=0\). For nonconstant solutions with \(y\ne1\), separate variables:
Rewrite the rational factor by expressing its numerator as \((y-1)+3\):
Integrating both sides gives
To check, differentiate the implicit relation with respect to \(x\):
Thus \(y'=\frac{y-1}{y+2}\), as required, wherever \(y\ne1\) and \(y\ne-2\). The equilibrium \(y=1\) was not part of the divided calculation, and \(y=-2\) remains excluded because the original equation is undefined there.
Worked Example: A Separable Initial-Value Problem
Worked Example: A Separable Initial-Value Problem
Solve \(y'=x\frac{y+1}{y-2}\) with \(y(0)=0\), expressing the answer implicitly.
The original equation is undefined at \(y=2\), and \(y=-1\) is an equilibrium because the numerator is zero there. Since the initial value is \(0\), the solution is not that equilibrium. For a non-equilibrium solution, separate variables:
Write \(y-2=(y+1)-3\), giving
Now integrate, keeping the constants together:
Apply \(y(0)=0\). The left side is \(0-3\ln 1=0\), and the right side is \(0+C\), so \(C=0\). The implicit solution is
For the branch through \(y=0\), we have \(y>-1\), so the logarithm may be written as \(\ln(y+1)\). Check the relation by differentiation:
This recovers the original equation wherever \(y\ne2\) and \(y\ne-1\). The initial value is also satisfied, since substituting \(x=0\) and \(y=0\) gives \(0=0\). The equilibrium \(y=-1\) is not represented by this separated calculation; it must be checked in the original equation.
Worked Example: A Repeated Linear Factor
Worked Example: A Repeated Linear Factor
Solve the equation \(y'=x\frac{y(y-2)^2}{y+1}\) implicitly for non-equilibrium solutions.
The original equation is undefined at \(y=-1\). Its numerator vanishes at \(y=0\) and \(y=2\), so \(y(x)=0\) and \(y(x)=2\) are equilibrium solutions. For non-equilibrium solutions, divide by \(y(y-2)^2\) and separate:
Because \(y\) and \(y-2\) are repeated linear factors in the denominator, use the form
Multiply by \(y(y-2)^2\):
Set \(y=0\): \(1=4A\), so \(A=\frac14\). Set \(y=2\): \(3=2C\), so \(C=\frac32\). The coefficient of \(y^2\) on the right is \(A+B\), which must equal \(0\); hence \(B=-\frac14\). Substituting these values verifies the remaining coefficient: \(-4A-2B+C=-1+\frac12+\frac32=1\), matching the coefficient of \(y\) on the left.
Integrate the resulting decomposition term by term:
For a check, differentiate the left side with respect to \(y\):
Thus differentiating the implicit relation with respect to \(x\) gives \(\frac{y+1}{y(y-2)^2}y'=x\), or \(y'=x\frac{y(y-2)^2}{y+1}\), which is the original equation. The derivation applies only where \(y\ne0\), \(y\ne2\), and \(y\ne-1\). The first two excluded values are the equilibria handled separately; the last is outside the domain of the equation.
Restrictions and a Common Sign Error
The domain of the original differential equation determines where a solution can be defined. If its right-hand side has a denominator that vanishes at \(y=a\), then the equation is undefined at \(y=a\); an implicit formula does not make that value admissible. Also check whether separation divided by a numerator factor that vanishes. Such values may yield equilibrium solutions and must be tested in the original equation.
A frequent integration error occurs with a squared linear factor. The incorrect calculation \(\int (y-2)^{-2}\,dy=(y-2)^{-1}+C\) has the wrong sign. Differentiating the proposed antiderivative gives \(\frac{d}{dy}(y-2)^{-1}=-(y-2)^{-2}\), not \(+(y-2)^{-2}\). The correct result is
Checking an antiderivative by differentiation catches this error immediately. It is equally important to differentiate the complete implicit relation and verify that it gives the original differential equation, rather than only a transformed equation with a factor divided out.
Check Your Understanding
For each question, keep track of the domain of the original equation as well as the algebra of integration.
- What terms are needed in a partial-fraction decomposition when the denominator contains \((y-a)^3\)?
- Decompose \(\frac{2y+1}{y(y+1)}\) into partial fractions.
- Why must a value at which the original differential equation is undefined remain excluded even if it appears in an implicit relation?
- For \(y'=y(y-1)\), which constant solutions might be lost by dividing by \(y(y-1)\)?
- Differentiate \(-\frac{1}{y+3}\) to check the sign of the antiderivative of \(\frac{1}{(y+3)^2}\).
