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Math101learn.math101.caSeparable Differential Equations
A rigorous method for separable equations, including equilibrium solutions, implicit forms, and maximal intervals.
Precise definition
An equation is separable if it can be written $y'=g(x)h(y)$. For non-equilibrium portions where $h(y)\ne0$, rearrange $dy/h(y)=g(x)dx$ and integrate. Any root $h(y_)=0$ gives a constant solution $y=y_$ that division would remove.
Notation and mathematical language
The differential notation abbreviates a justified chain-rule computation. If $H'(y)=1/h(y)$ and $G'(x)=g(x)$, then $d[H(y(x))]/dx=g(x)$, so $H(y)=G(x)+C$. The result may remain implicit.
Conceptual picture
Separation accumulates reciprocal state-dependent rate on one side and independent-variable effect on the other. Equilibria form barriers under uniqueness. An implicit formula can encode multiple branches, so initial data identify the relevant one.
Conditions and key results
Continuity assumptions support integration and local solution theory. Division requires excluding zeros temporarily; logarithms and roots create interval restrictions. Solving algebraically may introduce extraneous branches or miss finite-time blow-up.
A reliable strategy
- Factor the equation as $g(x)h(y)$ and find all zeros of $h$ first.
- On a non-equilibrium interval, separate variables and integrate both sides.
- Apply the initial condition before or after solving for $y$, retaining the correct branch.
- Restore equilibrium solutions and verify the original equation and maximal interval.
Fully worked example
Interpretation and application
Separable models include growth, decay, drag, and reaction kinetics. Exact separation solves the stated equation; it does not establish that the product-form rate law is causal or valid outside observed conditions.
Common mistakes
Verification and reasonableness
- Differentiate an implicit solution using the chain rule.
- Substitute the initial point and every restored equilibrium into the original equation.
- Use signs and equilibria to compare monotonicity with the explicit formula.
Practice
- Find equilibria of $y'=x(y-2)$.
- Separate $y'=xy$ for $y\ne0$.
- Why state a maximal interval?
Answers and brief solutions
- $y=2$.
- $dy/y=x\,dx$.
- An algebraic formula may encounter singularities that a solution cannot cross.
Further deduction
Definite integrals can encode initial data cleanly: $\int_{y_0}^{y(x)}du/h(u)=\int_{x_0}^xg(s)ds$. This form avoids merging unrelated constants and makes branch dependence visible. It is especially useful when the antiderivative cannot be inverted with elementary functions; an implicit exact solution is still a complete mathematical answer.
Separation can also prove qualitative facts. If $g(x)\ge0$ and $h(y)>0$ throughout a rectangle, then every solution segment there is nondecreasing even before integration. If $1/h(y)$ has a finite improper integral from $y_0$ to infinity while the accumulated $g$ reaches that value in finite time, the implicit formula predicts finite-time blow-up. Thus the separated integrals contain interval and growth information, not merely a path to an explicit formula.
In an autonomous equation $y'=h(y)$, phase-line arrows follow the sign of $h$. A zero where the sign changes from positive to negative is attracting; the reverse is repelling. A zero without a sign change can be semistable, approached from only one side.
Related topics
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
For $y'=2xy$ and $y(0)=3$, what is $y(1)$?
- $\ln|y|=x^2+C$, so $y=Ce^{x^2}$.
- $C=3$, hence $y(1)=3e$.
End of lesson
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