6.1 First-Order ODEs: Separable, Linear, and Exact Equations

6.1 First-Order ODEs: Separable, Linear, and Exact Equations

Ordinary differential equations (ODEs) constitute approximately 10% to 15% of the GRE Mathematics Subject Test. Mastery of first-order differential equations requires rapid classification by algebraic structure, rigorous tracking of domains and singular solutions, and fluency with integrating factors and exact differentials.

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Classification: Order, Linearity, and Normal Form

An ordinary differential equation relates an independent variable $x$, an unknown function $y(x)$, and its derivatives $y', y'', \dots, y^{(n)}$.

• Order: The order of an ODE is the order of the highest derivative appearing in the equation.
• Linearity: An $n$-th order ODE is linear if it can be written in the form: $a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \dots + a_1(x) y' + a_0(x) y = g(x)$ Crucially, linearity requires that $y$ and all its derivatives appear to the first power, are not multiplied together (e.g., $y y'$ is nonlinear), and do not appear inside nonlinear transcendental functions (e.g., $\sin y$, $e^y$, or $\sqrt{y'}$).
• Normal Form: A first-order ODE in normal form is expressed as $y' = f(x, y)$.

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Separable Differential Equations

A first-order ODE is separable if the derivative factors into an $x$-dependent function and a $y$-dependent function: $\frac{dy}{dx} = g(x) h(y)$

Analytical Solution Procedure

1. Identify Equilibrium (Singular) Solutions: Find all real roots $y_i$ such that $h(y_i) = 0$. The constant functions $y(x) \equiv y_i$ are solutions. Dividing by $h(y)$ assumes $h(y) \neq 0$, so omitting equilibrium solutions loses valid solutions.
2. Separate and Integrate: Rewrite the differential equation in separated form: $\frac{1}{h(y)}\,dy = g(x)\,dx \implies \int \frac{1}{h(y)}\,dy = \int g(x)\,dx + C$
3. Solve for Explicit Form: If feasible, isolate $y(x)$ and determine the constant $C$ from initial data.

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First-Order Linear Equations and Integrating Factors

A first-order linear ODE can always be written in the standard form: $y' + P(x)y = Q(x)$

If the leading coefficient $a_1(x) \neq 1$ in $a_1(x)y' + a_0(x)y = g(x)$, you must divide the entire equation by $a_1(x)$ before identifying $P(x)$ and $Q(x)$.

Derivation of the Integrating Factor

We seek a strictly positive multiplier $\mu(x)$ such that the left-hand side transforms into a single product derivative $\frac{d}{dx}[\mu(x)y]$: $\mu(x)y' + \mu(x)P(x)y = \frac{d}{dx}[\mu(x)y] = \mu(x)y' + \mu'(x)y$ Equating coefficients of $y$ yields the separable differential equation for $\mu(x)$: $\mu'(x) = \mu(x)P(x) \implies \frac{d\mu}{\mu} = P(x)\,dx \implies \mu(x) = \exp\left(\int P(x)\,dx\right)$

Multiplying the standard form by $\mu(x)$ gives: $\frac{d}{dx}[\mu(x)y] = \mu(x)Q(x)$ Integrating both sides yields the general solution formula: $y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)Q(x)\,dx + C\right)$

Interval of Validity: If $P(x)$ and $Q(x)$ are continuous on an open interval $I$ containing $x_0$, the unique solution to the IVP $y(x_0) = y_0$ exists and is differentiable across the entirety of $I$. Singularities in $P(x)$ define the boundary of the maximal existence interval.

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Exact Differential Equations and Potential Functions

A first-order equation written in total differential form is: $M(x, y)\,dx + N(x, y)\,dy = 0$

Condition for Exactness

The expression $M,dx + N,dy$ is the total differential $d\Psi = \frac{\partial \Psi}{\partial x},dx + \frac{\partial \Psi}{\partial y},dy$ of a scalar potential function $\Psi(x, y)$ if and only if mixed second partial derivatives commute (Clairaut-Schwarz Theorem). On a simply connected domain $D \subseteq \mathbb{R}^2$ where $M, N$ and their first partials are continuous: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ When this condition holds, the ODE is exact, and the general solution is implicitly defined by: $\Psi(x, y) = C$

Recovering the Potential Function $\Psi(x, y)$

1. Integrate $M(x, y)$ with respect to $x$, introducing an arbitrary function of integration $g(y)$: $\Psi(x, y) = \int M(x, y)\,dx + g(y)$
2. Differentiate $\Psi(x, y)$ with respect to $y$ and equate to $N(x, y)$: $\frac{\partial \Psi}{\partial y} = \frac{\partial}{\partial y}\left(\int M(x, y)\,dx\right) + g'(y) = N(x, y)$
3. Solve for $g'(y) = N(x, y) - \frac{\partial}{\partial y}\left(\int M(x, y),dx\right)$. Exactness guarantees that the right side is independent of $x$. Integrate to find $g(y)$.

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Integrating Factors for Non-Exact Equations

When $M_y \neq N_x$, we seek a multiplying function $\mu(x, y)$ such that $(\mu M),dx + (\mu N),dy = 0$ is exact: $\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N) \implies M\frac{\partial \mu}{\partial y} + \mu \frac{\partial M}{\partial y} = N\frac{\partial \mu}{\partial x} + \mu \frac{\partial N}{\partial x}$

On the GRE, testing for single-variable integrating factors is standard:

1. Function of $x$ alone, $\mu = \mu(x)$: Here $\frac{\partial \mu}{\partial y} = 0$ and $\frac{\partial \mu}{\partial x} = \mu'(x)$: $\frac{M_y - N_x}{N} = f(x) \implies \mu(x) = \exp\left(\int f(x)\,dx\right)$
2. Function of $y$ alone, $\mu = \mu(y)$: Here $\frac{\partial \mu}{\partial x} = 0$ and $\frac{\partial \mu}{\partial y} = \mu'(y)$: $\frac{N_x - M_y}{M} = g(y) \implies \mu(y) = \exp\left(\int g(y)\,dy\right)$

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Step-by-Step Worked Problems

Problem 1: Non-Exact Equation Requiring an Integrating Factor

Find the general implicit solution to $(3x y + y^2),dx + (x^2 + xy),dy = 0$.

Solution:

1. Identify $M(x, y) = 3xy + y^2$ and $N(x, y) = x^2 + xy$.
2. Compute partial derivatives: $M_y = 3x + 2y$ and $N_x = 2x + y$.
3. Check difference: $M_y - N_x = (3x + 2y) - (2x + y) = x + y \neq 0$ (not exact).
4. Evaluate single-variable criterion: $\frac{M_y - N_x}{N} = \frac{x + y}{x(x + y)} = \frac{1}{x}$ Since this depends solely on $x$, compute $\mu(x) = \exp\left(\int \frac{1}{x},dx\right) = x$.
5. Multiply the ODE by $\mu(x) = x$: $(3x^2 y + x y^2)\,dx + (x^3 + x^2 y)\,dy = 0$
6. Integrate $\tilde{M} = 3x^2 y + x y^2$ with respect to $x$: $\Psi(x, y) = x^3 y + \frac{1}{2} x^2 y^2 + g(y)$
7. Differentiate with respect to $y$: $\frac{\partial \Psi}{\partial y} = x^3 + x^2 y + g'(y) = x^3 + x^2 y \implies g'(y) = 0 \implies g(y) = C_0$
8. The general solution is $x^3 y + \frac{1}{2} x^2 y^2 = C$.

Problem 2: First-Order Linear Initial Value Problem

Solve $x y' + 2y = \sin x$ on $(0, \infty)$ subject to $y(\pi) = \frac{1}{\pi^2}$.

Solution:

1. Put into standard form by dividing by $x$: $y' + \frac{2}{x}y = \frac{\sin x}{x}$
2. Compute integrating factor: $\mu(x) = \exp\left(\int \frac{2}{x}\,dx\right) = e^{2\ln x} = x^2$
3. Multiply and integrate: $\frac{d}{dx}[x^2 y] = x^2 \left(\frac{\sin x}{x}\right) = x \sin x$ $x^2 y = \int x \sin x\,dx = -x \cos x + \sin x + C$
4. Isolate $y(x)$: $y(x) = -\frac{\cos x}{x} + \frac{\sin x}{x^2} + \frac{C}{x^2}$
5. Apply initial condition $y(\pi) = \frac{1}{\pi^2}$: $-\frac{\cos \pi}{\pi} + \frac{\sin \pi}{\pi^2} + \frac{C}{\pi^2} = \frac{1}{\pi} + \frac{C}{\pi^2} = \frac{1}{\pi^2} \implies C = 1 - \pi$
6. Thus $y(x) = -\frac{\cos x}{x} + \frac{\sin x + 1 - \pi}{x^2}$.

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GRE Exam Traps & Pitfalls

Trap 1: Calculating $\mu(x)$ Before Putting Linear ODE in Standard Form Given $x^2 y' + 3x y = 1$, students often mistakenly set $P(x) = 3x$ and write $\mu(x) = e^{3x^2/2}$. You must divide by the leading coefficient first: $y' + \frac{3}{x}y = \frac{1}{x^2}$, so $P(x) = \frac{3}{x}$ and $\mu(x) = x^3$.

Trap 2: Forgetting the Constant $C$ Before Dividing by $\mu(x)$ Writing $y(x) = \frac{1}{\mu(x)} \int \mu(x)Q(x),dx + C$ is wrong; the constant of integration belongs inside the parentheses, yielding $y(x) = \frac{1}{\mu(x)} \int \mu(x)Q(x),dx + \frac{C}{\mu(x)}$. Omitting this leads to incorrect initial condition evaluations.

Trap 3: Sign Reversal in the Non-Exact Integrating Factor Test Do not confuse $\frac{M_y - N_x}{N} = f(x)$ with $\frac{N_x - M_y}{M} = g(y)$. Reversing the signs produces incorrect integrating factors that fail to render the differential form exact.