Separable, Homogeneous and Linear First-Order Equations

Chapter 1 set up first-order equations and proved when they are well posed; this chapter supplies the three standard techniques that actually produce a closed-form solution: separation of variables, the homogeneous substitution $v=y/x$, and the integrating factor for linear equations. Each technique reduces the given ODE to one or two ordinary integrals, so the chapter is really about recognising which reduction applies and executing it correctly — including the bookkeeping steps (singular solutions, the constant of integration, the role of the integrating factor) that are the most common sources of lost marks.

Have to hand: techniques of integration from a first calculus course (substitution, partial fractions), and the vocabulary of Chapter 1 (order, linearity, IVP, equilibrium solution).

1. Separable Equations

A first-order ODE is separable if it can be written as

for some functions $g,h$ — the right side splits into a factor depending only on $x$ and a factor depending only on $y$.

Method. Wherever $h(y)\ne 0$, divide by $h(y)$ and integrate both sides with respect to $x$:

This is justified by the chain rule: if $H(y) = \int \frac{dy}{h(y)}$, then $\dfrac{d}{dx}H(y(x)) = H'(y)\,y' = \dfrac{y'}{h(y)} = g(x)$, and integrating both sides in $x$ recovers exactly the displayed equation.

Singular solutions. Every zero $y=y_0$ of $h(y)$ gives the constant function $y\equiv y_0$ as an extra solution, which the division step silently discards. These are exactly the equilibrium solutions of Chapter 1 (Section 4) and must be reported separately — a separable-equation answer that omits them is incomplete.

Worked example (logistic equation, solved explicitly). Solve $y' = y(1-y)$, $y(0)=y_0$ with $0<y_0<1$. Separating,

Partial fractions give $\dfrac{1}{y(1-y)} = \dfrac{1}{y} + \dfrac{1}{1-y}$, so the left side integrates to $\ln|y| - \ln|1-y| = \ln\left|\dfrac{y}{1-y}\right|$. Hence

Using $y(0)=y_0$ gives $A = y_0/(1-y_0)$, and solving for $y$,

As $x\to\infty$, $y(x)\to 1$ — matching the stability analysis of Chapter 1 (the equilibrium $y=1$ is asymptotically stable), and this is the "S-shaped" logistic curve. The discarded singular solutions $y\equiv 0$ and $y\equiv 1$ are precisely the two equilibria.

2. Homogeneous Equations

A first-order ODE is homogeneous (in the ODE sense — not to be confused with the "homogeneous" linear equations of Section 3, an unfortunate but standard clash of terminology) if it can be written as

for some function $f$ of a single variable, i.e. the right side depends on $x$ and $y$ only through the ratio $y/x$. Equivalently, $M(x,y)\,dx + N(x,y)\,dy=0$ is homogeneous in this sense if $M,N$ are homogeneous functions of the same degree ($M(tx,ty)=t^nM(x,y)$, likewise $N$), since then their ratio depends only on $y/x$.

Method (the substitution $v=y/x$). Set $y = vx$, so $v = y/x$ and, by the product rule, $y' = v + x\dfrac{dv}{dx}$. Substituting into $y'=f(y/x)=f(v)$ gives

which separates in $v$ and $x$:

This is why the technique belongs in this chapter: every homogeneous equation reduces to a separable one after exactly one substitution.

Worked example. Solve $\dfrac{dy}{dx} = \dfrac{x^2+y^2}{xy}$. Dividing numerator and denominator by $x^2$, the right side is $\dfrac{1+(y/x)^2}{y/x} = \dfrac{1+v^2}{v}$ with $v=y/x$, confirming the equation is homogeneous. Substituting $y=vx$, $y'=v+xv'$:

Separating, $v\,dv = \dfrac{dx}{x}$, so $\dfrac{v^2}{2} = \ln|x| + C$, i.e. $v^2 = 2\ln|x| + C_1$. Substituting back $v=y/x$:

3. Linear First-Order Equations

A first-order linear ODE, from Chapter 1, can always be arranged into standard form

Unlike general nonlinear first-order equations, this entire class has a single, universal method of solution.

Theorem (integrating factor). If $p,q$ are continuous on an interval $I$, define $\mu(x) = e^{\int p(x)\,dx}$ (any one antiderivative). Then on $I$,

and hence the general solution is

Proof. By the product rule, $\dfrac{d}{dx}[\mu y] = \mu y' + \mu' y$. Since $\mu' = p(x)\mu(x)$ (as $\mu=e^{\int p\,dx}$), this is $\mu y' + p\mu y = \mu(y'+py)$. If $y$ solves the ODE, $y'+py=q$, so $\dfrac{d}{dx}[\mu y] = \mu q$ exactly — the left side is now a single derivative, hence directly integrable. Conversely, integrating $\dfrac{d}{dx}[\mu y]=\mu q$ and dividing by $\mu$ (never zero, since it is an exponential) recovers every solution of the original equation. $\blacksquare$

The entire content of the method is the observation that $\mu(x)$ was engineered precisely so that $\mu y' + \mu' y$ collapses to the derivative of a product — the same combination that appears on the left after multiplying the standard form through by $\mu$.

Worked example (mixing problem, continued from Chapter 1). Solve $y' = 2 - \dfrac{y}{50}$, $y(0)=0$, i.e. $y' + \dfrac{1}{50}y = 2$ in standard form, so $p(x)=\tfrac{1}{50}$ (constant) and $q(x)=2$. The integrating factor is $\mu(x) = e^{\int (1/50)\,dx} = e^{x/50}$. Then

Applying $y(0)=0$: $0 = 100+C$, so $C=-100$, giving $y(x) = 100\bigl(1-e^{-x/50}\bigr)$ — exactly the formula asserted (and verified by direct substitution) in Chapter 1's exercises, now derived rather than guessed. Note the equilibrium value $y\to 100$ as $x\to\infty$, consistent with Chapter 1's stability analysis of the autonomous mixing equation.

4. Choosing the Right Method

Given an unfamiliar first-order equation, the practical order of checks is:

1. Is it already separable (right side factors as $g(x)h(y)$)? If so, separate directly — it is almost always the fastest method when it applies.
2. Is it linear in $y$, i.e. of the form $y'+p(x)y=q(x)$? If so, use the integrating factor; this always works for linear equations, with no guesswork.
3. Does the right side depend only on $y/x$ (homogeneous in the Section 2 sense)? If so, substitute $v=y/x$ to reduce to a separable equation in $v$ and $x$.

Many equations satisfy more than one of these criteria simultaneously (e.g. $y'=ky$ is both separable and linear), in which case either method works and typically gives the same effort.

5. Common Exam Pitfalls

• Dropping the singular solutions of a separable equation. Dividing by $h(y)$ is only valid where $h(y)\ne 0$; every zero of $h$ gives an extra constant solution that must be stated alongside the general solution.
• Substituting $v=y/x$ without replacing $y'$ correctly. The product rule gives $y' = v + x\,dv/dx$, not $y'=dv/dx$ — forgetting the $v$ term is the single most common algebra slip in this method.
• Forgetting that the "homogeneous" of Section 2 is a different concept from "homogeneous linear" (Chapters 4–5, where it means $q(x)\equiv 0$). The same word names two unrelated properties; state which sense is meant when it matters.
• Choosing the wrong constant of integration when computing $\mu(x)$. Any single antiderivative of $p(x)$ works in $\mu(x)=e^{\int p\,dx}$; adding an arbitrary constant there only rescales $\mu$ by a constant factor, which cancels out of the final formula, but students sometimes mistakenly carry an extra $+C$ through and corrupt the answer.
• Forgetting to divide by $\mu(x)$ at the end. The integrating-factor method produces $\mu y$ as an antiderivative of $\mu q$; the solution is $y = (\text{that antiderivative})/\mu(x)$, not the antiderivative itself.
• Misidentifying $p(x)$ and $q(x)$ when the equation is not yet in standard form. $y' = 3y + x^2$ must first be rearranged to $y'-3y=x^2$, giving $p(x)=-3$, not $p(x)=3$.

Summary

• Separable: $y'=g(x)h(y)$ reduces to $\int dy/h(y) = \int g(x)\,dx$; zeros of $h(y)$ give extra equilibrium (singular) solutions lost in the division step.
• Homogeneous ($y'=f(y/x)$): substitute $y=vx$, $y'=v+xv'$, reducing to the separable equation $x\,dv/dx = f(v)-v$.
• Linear ($y'+p(x)y=q(x)$): the integrating factor $\mu(x)=e^{\int p\,dx}$ makes $\dfrac{d}{dx}[\mu y]=\mu q$, giving the universal formula $y = \mu(x)^{-1}\left(\int \mu(x) q(x)\,dx + C\right)$.
• Method selection: check separability first, then linearity (always solvable via integrating factor), then the homogeneous ratio form.
• Worked results to remember: the logistic solution $y(x) = \left[1+\left(\tfrac{1-y_0}{y_0}\right)e^{-x}\right]^{-1}$, and the mixing-problem solution $y(x)=100(1-e^{-x/50})$ derived rigorously via the integrating factor.