From Separation to Integration
A separable differential equation can often be written as a product of a function of \(x\) and a function of \(y\). When those factors are rational expressions, separating the variables produces integrals that may involve polynomials, logarithms, or inverse tangent functions.
The main steps are familiar from earlier material on separating variables and using initial conditions: arrange all \(y\)-dependent factors with \(dy\), all \(x\)-dependent factors with \(dx\), integrate both sides, and use the given point to determine the constant. The details that matter are the domain of the rational expressions and the accuracy of each antiderivative.
For example, if \(dy/dx=g(x)h(y)\) and \(h(y)\ne 0\) on the part of the solution being considered, separation gives
The condition matters: dividing by \(h(y)\) is legitimate only where it is nonzero. In the examples below, we keep track of the expressions that must remain defined. A solution may be left in implicit form if solving explicitly for \(y\) would be difficult or unnecessary.
Why the Separation Method Works
Separation is a way to turn a differential equation into an equality of accumulated changes. The justification is the chain rule. Suppose integration produces a relation
where \(H'(y)=1/h(y)\) and \(G'(x)=g(x)\). Differentiating both sides with respect to \(x\) gives
Multiplying by \(h(y)\) recovers \(dy/dx=g(x)h(y)\). Thus, wherever the functions and derivatives are defined and the separation steps are valid, the integrated relation satisfies the original differential equation.
The reverse direction follows by separating the original equation and integrating on an interval where the integrands are defined. A solution curve through an initial point must therefore satisfy the integrated relation, with its constant determined by that point.
Worked Example: A Polynomial Antiderivative
Worked Example: A Polynomial Antiderivative
Solve \(dy/dx=x/(1+y^2)\) with \(y(0)=1\).
The denominator \(1+y^2\) is positive for every real \(y\), so the equation is defined for all real \(y\). Multiply by \(1+y^2\) and by \(dx\):
Integrate both sides:
Use the initial condition \(x=0,\ y=1\):
The particular solution is therefore described implicitly by
To check, differentiate the relation:
At \(x=0,\ y=1\), the implicit relation gives \(1+1/3=4/3\), so it also satisfies the initial condition. The implicit equation is a precise description of the solution curve; a simple explicit formula for \(y\) is not needed.
Rational Integrals Can Produce Inverse Tangents
A common rational expression in these equations is \(1/(1+u^2)\). Its antiderivative is \(\arctan u\), since the derivative of \(\arctan u\) is \(1/(1+u^2)\). The separated integrals need not both be polynomials.
Worked Example: An Inverse-Tangent Antiderivative
Solve \(dy/dx=1/[(1+x^2)(1+y^2)]\) with \(y(0)=0\).
Both denominator factors are positive for real inputs, so this equation is defined for every real \(x\) and \(y\). Multiply by \(1+y^2\) and by \(dx\):
Integrate:
Apply \(y(0)=0\), using \(\arctan 0=0\):
The solution curve is given implicitly by
Differentiate the equation:
This is the original equation, and substituting \(x=0,\ y=0\) into the implicit relation gives \(0=0\). Both checks hold. For each fixed \(x\), the left side \(y+y^3/3\) is strictly increasing in \(y\), because its derivative is \(1+y^2>0\). Consequently this relation specifies one real \(y\) for each real \(x\).
Logarithms and the Domain of a Solution
If a separated rational factor has the form \(1/(x+a)\), its antiderivative is \(\ln|x+a|\), on an interval that does not contain \(x=-a\). The absolute value is essential when the domain includes values on either side of the zero; any one solution interval must still avoid that point.
Worked Example: A Logarithmic Antiderivative
Solve \(dy/dx=(1+y^2)/(x+1)\) with \(y(0)=0\), and describe the interval containing \(x=0\) on which the resulting formula is defined.
The original equation is undefined at \(x=-1\). On an interval not containing \(-1\), separate:
Integrate the two rational expressions:
At \(x=0,\ y=0\), we have \(\arctan 0=0\) and \(\ln|0+1|=0\), so \(C=0\). Since the solution interval contains \(0\), it lies on the side \(x>-1\), where \(|x+1|=x+1\). Thus
The tangent formula is defined when its input is strictly between consecutive vertical asymptotes. For the branch containing \(x=0\), require
Therefore the interval containing \(0\) is \(e^{-\pi/2}-1<x<e^{\pi/2}-1\). Check by differentiating:
At \(x=0\), the formula gives \(y=\tan(0)=0\). It satisfies both the equation and the initial condition on the stated interval.
When an Explicit Branch Is Useful
Sometimes an integrated relation can be rearranged to give \(y\) explicitly. The initial condition then helps select the branch that passes through the required point. The equation and the initial value should both be checked after rearranging.
Worked Example: Selecting a Square-Root Branch
Solve \(dy/dx=x/(y+2)\) with \(y(0)=0\).
The differential equation is defined only where \(y\ne -2\). Multiply by \(y+2\) and by \(dx\), on an interval where this condition holds:
Integrating gives
The initial condition yields \(0=C\). Multiply by \(2\) and complete the square:
Taking square roots gives two algebraic branches. The condition \(y(0)=0\) requires \(y+2=2>0\) at \(x=0\), so the required branch is
For every real \(x\), \(\sqrt{x^2+4}>0\), so \(y+2=\sqrt{x^2+4}\ne0\); the branch remains within the domain of the original equation. Differentiate and compare with the right side:
The derivative matches the equation, and at \(x=0\) the explicit formula gives \(y=-2+2=0\). Both checks are complete.
A Common Integration Mistake
A frequent error is to separate correctly but integrate the \(y\)-side incorrectly. For the equation \(dy/dx=x/(1+y^2)\), separation gives \((1+y^2)\,dy=x\,dx\). A wrong calculation might claim
That is incorrect: differentiating \(y^2/2+y\) gives \(y+1\), not \(1+y^2\). The correct antiderivative is
A reliable check is to differentiate each antiderivative with respect to its own variable before using the initial condition. Then differentiate the final implicit relation with respect to \(x\) and confirm that it recovers the original differential equation.
A Practical Solution Process
For a separable equation with rational factors, use this sequence of checks:
Identify denominator zeros and other restrictions before rearranging.
Place all \(y\)-dependent factors with \(dy\), and all \(x\)-dependent factors with \(dx\), without dividing by an expression that could be zero on the solution being considered.
Use appropriate antiderivatives, including logarithms or inverse tangent where required.
Substitute the given point into the integrated relation to determine the constant.
Differentiate the relation or explicit formula, compare it with the equation, and verify the initial condition and domain.
The essential idea is that integration produces a relationship between \(x\) and \(y\), while the initial condition fixes which member of that family is intended. Rational factors make this process a useful application of antiderivatives; domain restrictions and accurate differentiation ensure that the resulting curve really solves the stated initial-value problem.
Check Your Understanding
For each question, keep track of both the integration and the domain of the resulting curve.
- Separate and integrate \(dy/dx=x/(1+y^2)\), and state the antiderivative on the \(y\)-side.
- For \(dy/dx=1/[(1+x^2)(1+y^2)]\) with \(y(0)=1\), find the implicit solution relation.
- Why must an interval of solution for \(dy/dx=(1+y^2)/(x+1)\) avoid \(x=-1\)?
- For \(dy/dx=x/(y+2)\) with \(y(0)=0\), which square-root branch satisfies the initial condition?
- What two checks should be made after using an initial condition to find a solution?
